https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Owjebra&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-04T02:25:23Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_14&diff=135473 2015 AIME II Problems/Problem 14 2020-10-21T01:36:10Z <p>Owjebra: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers satisfying &lt;math&gt;x^4y^5+y^4x^5=810&lt;/math&gt; and &lt;math&gt;x^3y^6+y^3x^6=945&lt;/math&gt;. Evaluate &lt;math&gt;2x^3+(xy)^3+2y^3&lt;/math&gt;.<br /> <br /> ==Solution==<br /> The expression we want to find is &lt;math&gt;2(x^3+y^3) + x^3y^3&lt;/math&gt;.<br /> <br /> Factor the given equations as &lt;math&gt;x^4y^4(x+y) = 810&lt;/math&gt; and &lt;math&gt;x^3y^3(x^3+y^3)=945&lt;/math&gt;, respectively. Dividing the latter by the former equation yields &lt;math&gt;\frac{x^2-xy+y^2}{xy} = \frac{945}{810}&lt;/math&gt;. Adding 3 to both sides and simplifying yields &lt;math&gt;\frac{(x+y)^2}{xy} = \frac{25}{6}&lt;/math&gt;. Solving for &lt;math&gt;x+y&lt;/math&gt; and substituting this expression into the first equation yields &lt;math&gt;\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810&lt;/math&gt;. Solving for &lt;math&gt;xy&lt;/math&gt;, we find that &lt;math&gt;xy = 3\sqrt{2}&lt;/math&gt;, so &lt;math&gt;x^3y^3 = 54&lt;/math&gt;. Substituting this into the second equation and solving for &lt;math&gt;x^3+y^3&lt;/math&gt; yields &lt;math&gt;x^3+y^3=\frac{35}{2}&lt;/math&gt;. So, the expression to evaluate is equal to &lt;math&gt;2 \times \frac{35}{2} + 54 = \boxed{089}&lt;/math&gt;.<br /> <br /> Note that since the value we want to find is &lt;math&gt;2(x^3+y^3)+x^3y^3&lt;/math&gt;, we can convert &lt;math&gt;2(x^3+y^3)&lt;/math&gt; into an expression in terms of &lt;math&gt;x^3y^3&lt;/math&gt;, since from the second equation which is &lt;math&gt;x^3y^3(x^3+y^3)=945&lt;/math&gt;, we see that &lt;math&gt;2(x^3+y^3)=1890+x^6y^y,&lt;/math&gt; and thus the value is &lt;math&gt;\frac{1890+x^6y^6}{x^3y^3}.&lt;/math&gt; Since we've already found &lt;math&gt;x^3y^3,&lt;/math&gt; we substitute and find the answer to be 89.<br /> <br /> ==Solution 2==<br /> Factor the given equations as &lt;math&gt;x^4y^4(x+y) = 810&lt;/math&gt; and &lt;math&gt;x^3y^3(x+y)(x^2-xy+y^2)=945&lt;/math&gt;, respectively. By the first equation, &lt;math&gt;x+y=\frac{810}{x^4y^4}&lt;/math&gt;. Plugging this in to the second equation and simplifying yields &lt;math&gt;(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}&lt;/math&gt;. Now substitute &lt;math&gt;\frac{x}{y}=a&lt;/math&gt;. Solving the quadratic in &lt;math&gt;a&lt;/math&gt;, we get &lt;math&gt;a=\frac{x}{y}=\frac{2}{3}&lt;/math&gt; or &lt;math&gt;\frac{3}{2}&lt;/math&gt; As both of the original equations were symmetric in &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, WLOG, let &lt;math&gt;\frac{x}{y}=\frac{2}{3}&lt;/math&gt;, so &lt;math&gt;x=\frac{2}{3}y&lt;/math&gt;. Now plugging this in to either one of the equations, we get the solutions &lt;math&gt;y=\frac{3(2^{\frac{2}{3}})}{2}&lt;/math&gt;, &lt;math&gt;x=2^{\frac{2}{3}}&lt;/math&gt;. Now plugging into what we want, we get &lt;math&gt;8+54+27=\boxed{089}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Add three times the first equation to the second equation and factor to get &lt;math&gt;(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375&lt;/math&gt;. Taking the cube root yields &lt;math&gt;xy(x+y)=15&lt;/math&gt;. Noting that the first equation is &lt;math&gt;(xy)^3\cdot(xy(x+y))=810&lt;/math&gt;, we find that &lt;math&gt;(xy)^3=\frac{810}{15}=54&lt;/math&gt;. Plugging this into the second equation and dividing yields &lt;math&gt;x^3+y^3 = \frac{945}{54} = \frac{35}{2}&lt;/math&gt;. Thus the sum required, as noted in Solution 1, is &lt;math&gt;54+\frac{35}{2}\cdot2 = \boxed{089}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> As with the other solutions, factor. But this time, let &lt;math&gt;a=xy&lt;/math&gt; and &lt;math&gt;b=x+y&lt;/math&gt;. Then &lt;math&gt;a^4b=810&lt;/math&gt;. Notice that &lt;math&gt;x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)&lt;/math&gt;. Now, if we divide the second equation by the first one, we get &lt;math&gt;7/6 = \frac{b^2-3a}{a}&lt;/math&gt;; then &lt;math&gt;\frac{b^2}{a}=\frac{25}{6}&lt;/math&gt;. Therefore, &lt;math&gt;a = \frac{6}{25}b^2&lt;/math&gt;. Substituting &lt;math&gt;a&lt;/math&gt; into &lt;math&gt;b&lt;/math&gt; in equation 2 gives us &lt;math&gt;b^3 = \frac{5^3}{2}&lt;/math&gt;; we are looking for &lt;math&gt;2b(b^2-3a)+a^3&lt;/math&gt;. Finding &lt;math&gt;a&lt;/math&gt;, we get &lt;math&gt;35&lt;/math&gt;. Substituting into the first equation, we get &lt;math&gt;b=54&lt;/math&gt;. Our final answer is &lt;math&gt;35+54=\boxed{089}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Owjebra https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_9&diff=132381 2017 AIME I Problems/Problem 9 2020-08-23T02:17:58Z <p>Owjebra: /* Solution 3 */ $9 \mid n$ was confused with $n \mid 9$, and similarly with $n + 1$.</p> <hr /> <div>==Problem 9==<br /> Let &lt;math&gt;a_{10} = 10&lt;/math&gt;, and for each integer &lt;math&gt;n &gt;10&lt;/math&gt; let &lt;math&gt;a_n = 100a_{n - 1} + n&lt;/math&gt;. Find the least &lt;math&gt;n &gt; 10&lt;/math&gt; such that &lt;math&gt;a_n&lt;/math&gt; is a multiple of &lt;math&gt;99&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Writing out the recursive statement for &lt;math&gt;a_n, a_{n-1}, \dots, a_{10}&lt;/math&gt; and summing them gives &lt;cmath&gt;a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10&lt;/cmath&gt;<br /> Which simplifies to &lt;cmath&gt;a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;a_n&lt;/math&gt; is divisible by 99 if and only if &lt;math&gt;\frac{1}{2}(n+10)(n-9)&lt;/math&gt; is divisible by 99, so &lt;math&gt;(n+10)(n-9)&lt;/math&gt; needs to be divisible by 9 and 11. Assume that &lt;math&gt;n+10&lt;/math&gt; is a multiple of 11. Writing out a few terms, &lt;math&gt;n=12, 23, 34, 45&lt;/math&gt;, we see that &lt;math&gt;n=45&lt;/math&gt; is the smallest &lt;math&gt;n&lt;/math&gt; that works in this case. Next, assume that &lt;math&gt;n-9&lt;/math&gt; is a multiple of 11. Writing out a few terms, &lt;math&gt;n=20, 31, 42, 53&lt;/math&gt;, we see that &lt;math&gt;n=53&lt;/math&gt; is the smallest &lt;math&gt;n&lt;/math&gt; that works in this case. The smallest &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> Note that we can also construct the solution using CRT by assuming either &lt;math&gt;11&lt;/math&gt; divides &lt;math&gt;n+10&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt; divides &lt;math&gt;n-9&lt;/math&gt;, or &lt;math&gt;9&lt;/math&gt; divides &lt;math&gt;n+10&lt;/math&gt; and &lt;math&gt;11&lt;/math&gt; divides &lt;math&gt;n-9&lt;/math&gt;, and taking the smaller solution.<br /> <br /> ==Solution 2==<br /> &lt;cmath&gt;a_n \equiv a_{n-1} + n \pmod {99} &lt;/cmath&gt;<br /> By looking at the first few terms, we can see that <br /> &lt;cmath&gt;a_n \equiv 10+11+12+ \dots + n \pmod {99} &lt;/cmath&gt;<br /> This implies<br /> &lt;cmath&gt;a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} &lt;/cmath&gt;<br /> Since &lt;math&gt;a_n \equiv 0 \pmod {99}&lt;/math&gt;, we can rewrite the equivalence, and simplify <br /> &lt;cmath&gt;0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv n(n+1) - 90 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv 4n^2+4n+36 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv (2n+1)^2+35 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;64 \equiv (2n+1)^2 \pmod {99} &lt;/cmath&gt;<br /> The only squares that are congruent to &lt;math&gt;64 \pmod {99}&lt;/math&gt; are &lt;math&gt;(\pm 8)^2&lt;/math&gt; and &lt;math&gt;(\pm 19)^2&lt;/math&gt;, so <br /> &lt;cmath&gt;2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}&lt;/cmath&gt;<br /> &lt;math&gt;2n+1 \equiv -8 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=45&lt;/math&gt; as the smallest integer solution.<br /> <br /> &lt;math&gt;2n+1 \equiv 8 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=53&lt;/math&gt; as the smallest integer solution.<br /> <br /> &lt;math&gt;2n+1 \equiv -19 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=89&lt;/math&gt; as the smallest integer solution.<br /> <br /> &lt;math&gt;2n+1 \equiv 19 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=9&lt;/math&gt; as the smallest integer solution. However, &lt;math&gt;n&lt;/math&gt; must be greater than &lt;math&gt;10&lt;/math&gt;.<br /> <br /> The smallest positive integer solution greater than &lt;math&gt;10&lt;/math&gt; is &lt;math&gt;n=\boxed{045}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> &lt;math&gt;a_n=a_{n-1} + n \pmod{99}&lt;/math&gt;. Using the steps of the previous solution we get up to &lt;math&gt;n^2+n \equiv 90 \pmod{99}&lt;/math&gt;. This gives away the fact that &lt;math&gt;(n)(n+1) \equiv 0 \pmod{9} \implies n \equiv \{0, 8\} \pmod{9}&lt;/math&gt; so either &lt;math&gt;n&lt;/math&gt; or &lt;math&gt;n+1&lt;/math&gt; must be a multiple of 9. <br /> <br /> Case 1 (&lt;math&gt;9 \mid n&lt;/math&gt;): Say &lt;math&gt;n=9x&lt;/math&gt; and after simplification &lt;math&gt;x(9x+1) = 10 \pmod{90} \forall x \in \mathbb{Z}&lt;/math&gt;. <br /> <br /> Case 2: (&lt;math&gt;9 \mid n+1&lt;/math&gt;): Say &lt;math&gt;n=9a-1&lt;/math&gt; and after simplification &lt;math&gt;(9a-1)(a) = 10 \pmod{90} \forall a \in \mathbb{Z}&lt;/math&gt;.<br /> <br /> As a result &lt;math&gt;a&lt;/math&gt; must be a divisor of &lt;math&gt;10&lt;/math&gt; and after doing some testing in both cases the smallest value that works is &lt;math&gt;x=5 \implies \boxed{045}&lt;/math&gt;.<br /> <br /> ~First<br /> <br /> ==Solution 4 (not good, risky)==<br /> We just notice that &lt;math&gt;100 \equiv 1 \pmod{99}&lt;/math&gt;, so we are just trying to find &lt;math&gt;10 + 11 + 12 + \cdots + n&lt;/math&gt; modulo &lt;math&gt;99&lt;/math&gt;, or &lt;math&gt;\dfrac{n(n+1)}{2} - 45&lt;/math&gt; modulo &lt;math&gt;99&lt;/math&gt;. Also, the sum to &lt;math&gt;44&lt;/math&gt; is divisible by &lt;math&gt;99&lt;/math&gt;, and is the first one that is. Thus, if we sum to &lt;math&gt;45&lt;/math&gt; the &lt;math&gt;45&lt;/math&gt; is cut off and thus is just a sum to &lt;math&gt;44&lt;/math&gt;.<br /> <br /> Without checking whether there are other sums congruent to &lt;math&gt;45 \pmod{99}&lt;/math&gt;, we can just write the answer to be &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt;b_n = 2a_{n+10}&lt;/math&gt;. We can find a formula for &lt;math&gt;b_n&lt;/math&gt;: <br /> <br /> &lt;math&gt;b_n = (20+n)(n+1)&lt;/math&gt;. <br /> <br /> Notice that both can't have a factor of 3. Thus we can limit our search range of n to &lt;math&gt;n \equiv 7,8 \pmod{9}&lt;/math&gt;. Testing values for n in our search range (like 7,8,16,17,25,26...), we get that 35 is the least n. But, don't write that down! Remember, &lt;math&gt;b_n = 2a_{n+10}&lt;/math&gt;, so, the 35th term in b corresponds to the 45th term in a. Thus our answer is &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> -AlexLikeMath<br /> <br /> ==Solution 6 (bash, slower, but safer)==<br /> <br /> The first thing you should realize is that each term after the tenth is another two-digit number chained to the last number. &lt;math&gt;10, 1011, 101112, \dots&lt;/math&gt;. Now the fact that the sequence starts at &lt;math&gt;10&lt;/math&gt; can be completely discarded for this solution. Just consider &lt;math&gt;a(10)&lt;/math&gt; then same as &lt;math&gt;a(1)&lt;/math&gt;, and we can add nine to the answer at the end.<br /> <br /> <br /> The second step is to split &lt;math&gt;99&lt;/math&gt; as &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;11&lt;/math&gt; and solve for divisibility rules individually. Let's start with &lt;math&gt;11&lt;/math&gt; because it gives us the most information to continue.<br /> <br /> <br /> In any number generated, if the numbers don't go beyond &lt;math&gt;20&lt;/math&gt;, then the highest number we can get is &lt;math&gt;10111213141516171819&lt;/math&gt;, with every odd digit being &lt;math&gt;1&lt;/math&gt;. This is a little risky because we are assuming that it doesn't exceed &lt;math&gt;20&lt;/math&gt;. If someone wanted to be absolutely sure they could continue, but this is unnecessary later and a big hassle. Anyways, now we write an equation to check for divisibility by &lt;math&gt;11&lt;/math&gt;. The expression being &lt;math&gt;\frac{((n-1) + 0)\cdot n)}{2}-n&lt;/math&gt;.<br /> <br /> The concept here is to add &lt;math&gt;0&lt;/math&gt; to the &lt;math&gt;n-1^{th}&lt;/math&gt; term altogether, then subtract the number of ones in it, which is &lt;math&gt;n&lt;/math&gt;. Simplify to &lt;math&gt;\frac{n(n-3)}{2}&lt;/math&gt; congruent to &lt;math&gt;0 \pmod{11}&lt;/math&gt;. Now notice the divide by two can be discarded because one of &lt;math&gt;n&lt;/math&gt; or &lt;math&gt;(n-3)&lt;/math&gt; will be even. So if &lt;math&gt;n&lt;/math&gt; or &lt;math&gt;n-3&lt;/math&gt; is to be divisible by &lt;math&gt;11&lt;/math&gt;, we can make a simple list. <br /> <br /> &lt;cmath&gt;n = 0, 3, 11, 14, 22, 25, 33, 36, 44, 47, \dots&lt;/cmath&gt;<br /> <br /> Now we test each &lt;math&gt;n&lt;/math&gt; for divisibility by &lt;math&gt;9&lt;/math&gt;.<br /> This is done by making a list that ultimately calculates the sum of every digit in the large number.<br /> &lt;math&gt;n(1)&lt;/math&gt; to &lt;math&gt;n(10)&lt;/math&gt; has the first digit &lt;math&gt;1&lt;/math&gt;. &lt;math&gt;n(11)&lt;/math&gt; to &lt;math&gt;n(20)&lt;/math&gt; has the first digit &lt;math&gt;2&lt;/math&gt;, and so on.<br /> The necessary thing to realize is that the sum of all digits &lt;math&gt;0-9&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;, so we only have to solve for the sum of the first digits, and then the short list of second digits.<br /> <br /> For example, let's test &lt;math&gt;n=25&lt;/math&gt;.<br /> <br /> So we know that &lt;math&gt;25&lt;/math&gt; include both &lt;math&gt;1-10&lt;/math&gt; and &lt;math&gt;11-20&lt;/math&gt;, so that's &lt;math&gt;10 + 20&lt;/math&gt; right away. &lt;math&gt;21-25&lt;/math&gt; contains &lt;math&gt;5&lt;/math&gt; numbers that have the first digit &lt;math&gt;3&lt;/math&gt;, so &lt;math&gt;+15&lt;/math&gt;. Then we add &lt;math&gt;0-4&lt;/math&gt; together, which is &lt;math&gt;10&lt;/math&gt;. &lt;math&gt;10+20+15+10=55&lt;/math&gt;, which is not divisible by &lt;math&gt;9&lt;/math&gt;, so it is not the answer. <br /> <br /> Do this for just a minute you get that &lt;math&gt;36&lt;/math&gt; sums to &lt;math&gt;99&lt;/math&gt;, a multiple of nine! So &lt;math&gt;n(36)&lt;/math&gt; is the answer, right? Don't forget we have to add &lt;math&gt;9&lt;/math&gt; because we translated &lt;math&gt;n(10)&lt;/math&gt; to &lt;math&gt;n(1)&lt;/math&gt; at the very beginning!<br /> Finally, after a short bash, we get &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> (LaTeX by PureSwag)<br /> <br /> ==See also==<br /> {{AIME box|year=2017|n=I|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Owjebra https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_6&diff=130334 2011 AIME II Problems/Problem 6 2020-08-03T18:47:24Z <p>Owjebra: /* Solution 4 (quick) */</p> <hr /> <div>==Problem 6==<br /> <br /> Define an ordered quadruple of integers &lt;math&gt;(a, b, c, d)&lt;/math&gt; as interesting if &lt;math&gt;1 \le a&lt;b&lt;c&lt;d \le 10&lt;/math&gt;, and &lt;math&gt;a+d&gt;b+c&lt;/math&gt;. How many interesting ordered quadruples are there?<br /> <br /> ==Solution 1==<br /> Rearranging the [[inequality]] we get &lt;math&gt;d-c &gt; b-a&lt;/math&gt;. Let &lt;math&gt;e = 11&lt;/math&gt;, then &lt;math&gt;(a, b-a, c-b, d-c, e-d)&lt;/math&gt; is a partition of 11 into 5 positive integers or equivalently:<br /> &lt;math&gt;(a-1, b-a-1, c-b-1, d-c-1, e-d-1)&lt;/math&gt; is a [[partition]] of 6 into 5 non-negative integer parts. Via a standard stars and bars argument, the number of ways to partition 6 into 5 non-negative parts is &lt;math&gt;\binom{6+4}4 = \binom{10}4 = 210&lt;/math&gt;. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry, there are as many partitions where the fourth is less than the second. So, if &lt;math&gt;N&lt;/math&gt; is the number of partitions where the second element is equal to the fourth, our answer is &lt;math&gt;(210-N)/2&lt;/math&gt;.<br /> <br /> We find &lt;math&gt;N&lt;/math&gt; as a sum of 4 cases:<br /> * two parts equal to zero, &lt;math&gt;\binom82 = 28&lt;/math&gt; ways,<br /> * two parts equal to one, &lt;math&gt;\binom62 = 15&lt;/math&gt; ways,<br /> * two parts equal to two, &lt;math&gt;\binom42 = 6&lt;/math&gt; ways,<br /> * two parts equal to three, &lt;math&gt;\binom22 = 1&lt;/math&gt; way.<br /> Therefore, &lt;math&gt;N = 28 + 15 + 6 + 1 = 50&lt;/math&gt; and our answer is &lt;math&gt;(210 - 50)/2 = \fbox{080}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let us consider our quadruple (a,b,c,d) as the following image xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding &lt;math&gt;\binom83 = 56&lt;/math&gt; ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding &lt;math&gt;\binom 63 = 20&lt;/math&gt; ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding &lt;math&gt;\binom43 = 4&lt;/math&gt; ways and there cannot be 3 holders between a and b so our total is 56+20+4=&lt;math&gt;\fbox{080}&lt;/math&gt;.<br /> <br /> ==Solution 3 (&lt;i&gt;Slightly&lt;/i&gt; bashy)==<br /> We first start out when the value of &lt;math&gt;a=1&lt;/math&gt;. <br /> <br /> Doing casework, we discover that &lt;math&gt;d=5,6,7,8,9,10&lt;/math&gt;. We quickly find a pattern.<br /> <br /> Now, doing this for the rest of the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, we see that the answer is simply:<br /> <br /> <br /> &lt;math&gt;(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)&lt;/math&gt;<br /> &lt;math&gt;+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\boxed{080}&lt;/math&gt;<br /> <br /> ==Solution 4 (quick)==<br /> <br /> Notice that if &lt;math&gt;a+d&gt;b+c&lt;/math&gt;, then &lt;math&gt;(11-a)+(11-d)&lt;(11-b)+(11-c)&lt;/math&gt;, so there is a 1-to-1 correspondence between the number of ordered quadruples with &lt;math&gt;a+d&gt;b+c&lt;/math&gt; and the number of ordered quadruples with &lt;math&gt;a+d&lt;b+c&lt;/math&gt;. <br /> <br /> Quick counting gives that the number of ordered quadruples with &lt;math&gt;a+d=b+c&lt;/math&gt; is 50. To count this, consider our numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 8, 9, 10&lt;/math&gt;. Notice that if, for example, &lt;math&gt;a+d=b+c=8&lt;/math&gt;, that the average of &lt;math&gt;a,d&lt;/math&gt; and &lt;math&gt;b,c&lt;/math&gt; must both be &lt;math&gt;4&lt;/math&gt;. In this way, there is a symmetry for this case, centered at &lt;math&gt;4&lt;/math&gt;. If instead, say, &lt;math&gt;a+d=b+c=7&lt;/math&gt;, an odd number, then there is symmetry with &lt;math&gt;(a,d);(b,c)&lt;/math&gt; about &lt;math&gt;3.5&lt;/math&gt;. Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at &lt;math&gt;2.5,3,8,8.5&lt;/math&gt;, there is &lt;math&gt;1&lt;/math&gt; case for each and so on, until centered at &lt;math&gt;5.5&lt;/math&gt;, there are &lt;math&gt;10&lt;/math&gt; possible cases. Adding these all, we have &lt;math&gt;2(1+3+6)+10=50&lt;/math&gt;.<br /> <br /> Thus the answer is &lt;math&gt;\frac{\binom{10}{4}-50}{2} = \boxed{080}.&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> <br /> Think about a,b,c,and d as distinct objects, that we must place in 4 of 10 spaces. However, in only 1 of 24 of these combinations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is &lt;math&gt;(10*9*8*7/24)=210&lt;/math&gt; <br /> <br /> Next, intuitively, the number of quadruples where &lt;math&gt;a+d&gt;b+c&lt;/math&gt; is equal to the number of quadruples where &lt;math&gt;a+d&lt;b+c&lt;/math&gt;. So we need to find the number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when &lt;math&gt;a-d&lt;/math&gt; ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where &lt;math&gt;a+d=b+c&lt;/math&gt;. <br /> <br /> So the answer will be &lt;math&gt;\frac{210-50}{2} = \boxed{080}.&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Owjebra https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_6&diff=130333 2011 AIME II Problems/Problem 6 2020-08-03T18:46:44Z <p>Owjebra: /* Solution 4 (quick) */ Explanation for counting $a+d=b+c$.</p> <hr /> <div>==Problem 6==<br /> <br /> Define an ordered quadruple of integers &lt;math&gt;(a, b, c, d)&lt;/math&gt; as interesting if &lt;math&gt;1 \le a&lt;b&lt;c&lt;d \le 10&lt;/math&gt;, and &lt;math&gt;a+d&gt;b+c&lt;/math&gt;. How many interesting ordered quadruples are there?<br /> <br /> ==Solution 1==<br /> Rearranging the [[inequality]] we get &lt;math&gt;d-c &gt; b-a&lt;/math&gt;. Let &lt;math&gt;e = 11&lt;/math&gt;, then &lt;math&gt;(a, b-a, c-b, d-c, e-d)&lt;/math&gt; is a partition of 11 into 5 positive integers or equivalently:<br /> &lt;math&gt;(a-1, b-a-1, c-b-1, d-c-1, e-d-1)&lt;/math&gt; is a [[partition]] of 6 into 5 non-negative integer parts. Via a standard stars and bars argument, the number of ways to partition 6 into 5 non-negative parts is &lt;math&gt;\binom{6+4}4 = \binom{10}4 = 210&lt;/math&gt;. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry, there are as many partitions where the fourth is less than the second. So, if &lt;math&gt;N&lt;/math&gt; is the number of partitions where the second element is equal to the fourth, our answer is &lt;math&gt;(210-N)/2&lt;/math&gt;.<br /> <br /> We find &lt;math&gt;N&lt;/math&gt; as a sum of 4 cases:<br /> * two parts equal to zero, &lt;math&gt;\binom82 = 28&lt;/math&gt; ways,<br /> * two parts equal to one, &lt;math&gt;\binom62 = 15&lt;/math&gt; ways,<br /> * two parts equal to two, &lt;math&gt;\binom42 = 6&lt;/math&gt; ways,<br /> * two parts equal to three, &lt;math&gt;\binom22 = 1&lt;/math&gt; way.<br /> Therefore, &lt;math&gt;N = 28 + 15 + 6 + 1 = 50&lt;/math&gt; and our answer is &lt;math&gt;(210 - 50)/2 = \fbox{080}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let us consider our quadruple (a,b,c,d) as the following image xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding &lt;math&gt;\binom83 = 56&lt;/math&gt; ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding &lt;math&gt;\binom 63 = 20&lt;/math&gt; ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding &lt;math&gt;\binom43 = 4&lt;/math&gt; ways and there cannot be 3 holders between a and b so our total is 56+20+4=&lt;math&gt;\fbox{080}&lt;/math&gt;.<br /> <br /> ==Solution 3 (&lt;i&gt;Slightly&lt;/i&gt; bashy)==<br /> We first start out when the value of &lt;math&gt;a=1&lt;/math&gt;. <br /> <br /> Doing casework, we discover that &lt;math&gt;d=5,6,7,8,9,10&lt;/math&gt;. We quickly find a pattern.<br /> <br /> Now, doing this for the rest of the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, we see that the answer is simply:<br /> <br /> <br /> &lt;math&gt;(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)&lt;/math&gt;<br /> &lt;math&gt;+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\boxed{080}&lt;/math&gt;<br /> <br /> ==Solution 4 (quick)==<br /> <br /> Notice that if &lt;math&gt;a+d&gt;b+c&lt;/math&gt;, then &lt;math&gt;(11-a)+(11-d)&lt;(11-b)+(11-c)&lt;/math&gt;, so there is a 1-to-1 correspondence between the number of ordered quadruples with &lt;math&gt;a+d&gt;b+c&lt;/math&gt; and the number of ordered quadruples with &lt;math&gt;a+d&lt;b+c&lt;/math&gt;. <br /> <br /> Quick counting gives that the number of ordered quadruples with &lt;math&gt;a+d=b+c&lt;/math&gt; is 50. To count this, consider our numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 8, 9 10&lt;/math&gt;. Notice that if, for example, &lt;math&gt;a+d=b+c=8&lt;/math&gt;, that the average of &lt;math&gt;a,d&lt;/math&gt; and &lt;math&gt;b,c&lt;/math&gt; must both be &lt;math&gt;4&lt;/math&gt;. In this way, there is a symmetry for this case, centered at &lt;math&gt;4&lt;/math&gt;. If instead, say, &lt;math&gt;a+d=b+c=7&lt;/math&gt;, an odd number, then there is symmetry with &lt;math&gt;(a,d);(b,c)&lt;/math&gt; about &lt;math&gt;3.5&lt;/math&gt;. Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at &lt;math&gt;2.5,3,8,8.5&lt;/math&gt;, there is &lt;math&gt;1&lt;/math&gt; case for each and so on, until centered at &lt;math&gt;5.5&lt;/math&gt;, there are &lt;math&gt;10&lt;/math&gt; possible cases. Adding these all, we have &lt;math&gt;2(1+3+6)+10=50&lt;/math&gt;.<br /> <br /> Thus the answer is &lt;math&gt;\frac{\binom{10}{4}-50}{2} = \boxed{080}.&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> <br /> Think about a,b,c,and d as distinct objects, that we must place in 4 of 10 spaces. However, in only 1 of 24 of these combinations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is &lt;math&gt;(10*9*8*7/24)=210&lt;/math&gt; <br /> <br /> Next, intuitively, the number of quadruples where &lt;math&gt;a+d&gt;b+c&lt;/math&gt; is equal to the number of quadruples where &lt;math&gt;a+d&lt;b+c&lt;/math&gt;. So we need to find the number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when &lt;math&gt;a-d&lt;/math&gt; ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where &lt;math&gt;a+d=b+c&lt;/math&gt;. <br /> <br /> So the answer will be &lt;math&gt;\frac{210-50}{2} = \boxed{080}.&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Owjebra https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_25&diff=128592 2011 AMC 12A Problems/Problem 25 2020-07-18T19:45:30Z <p>Owjebra: /* Solution */</p> <hr /> <div>== Problem ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;\angle BAC = 60^{\circ}&lt;/math&gt;, &lt;math&gt;\angle CBA \leq 90^{\circ}&lt;/math&gt;, &lt;math&gt;BC=1&lt;/math&gt;, and &lt;math&gt;AC \geq AB&lt;/math&gt;. Let &lt;math&gt;H&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, and &lt;math&gt;O&lt;/math&gt; be the orthocenter, incenter, and circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, respectively. Assume that the area of pentagon &lt;math&gt;BCOIH&lt;/math&gt; is the maximum possible. What is &lt;math&gt;\angle CBA&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 60^{\circ} \qquad<br /> \textbf{(B)}\ 72^{\circ} \qquad<br /> \textbf{(C)}\ 75^{\circ} \qquad<br /> \textbf{(D)}\ 80^{\circ} \qquad<br /> \textbf{(E)}\ 90^{\circ} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;\angle CAB=A&lt;/math&gt;, &lt;math&gt;\angle ABC=B&lt;/math&gt;, &lt;math&gt;\angle BCA=C&lt;/math&gt; for convenience.<br /> <br /> It's well-known that &lt;math&gt;\angle BOC=2A&lt;/math&gt;, &lt;math&gt;\angle BIC=90+\frac{A}{2}&lt;/math&gt;, and &lt;math&gt;\angle BHC=180-A&lt;/math&gt; (verifiable by angle chasing). Then, as &lt;math&gt;A=60&lt;/math&gt;, it follows that &lt;math&gt;\angle BOC=\angle BIC=\angle BHC=120&lt;/math&gt; and consequently pentagon &lt;math&gt;BCOIH&lt;/math&gt; is cyclic. Observe that &lt;math&gt;BC=1&lt;/math&gt; is fixed, hence the circumcircle of cyclic pentagon &lt;math&gt;BCOIH&lt;/math&gt; is also fixed. Similarly, as &lt;math&gt;OB=OC&lt;/math&gt;(both are radii), it follows that &lt;math&gt;O&lt;/math&gt; and also &lt;math&gt;[BCO]&lt;/math&gt; is fixed. Since &lt;math&gt;[BCOIH]=[BCO]+[BOIH]&lt;/math&gt; is maximal, it suffices to maximize &lt;math&gt;[BOIH]&lt;/math&gt;.<br /> <br /> Verify that &lt;math&gt;\angle IBC=\frac{B}{2}&lt;/math&gt;, &lt;math&gt;\angle HBC=90-C&lt;/math&gt; by angle chasing; it follows that &lt;math&gt;\angle IBH=\angle HBC-\angle IBC=90-C-\frac{B}{2}=\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}&lt;/math&gt; since &lt;math&gt;A+B+C=180\implies\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90&lt;/math&gt; by Triangle Angle Sum. Similarly, &lt;math&gt;\angle OBC=(180-120)/2=30&lt;/math&gt; (isosceles base angles are equal), hence &lt;cmath&gt;\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}&lt;/cmath&gt; <br /> Since &lt;math&gt;\angle IBH=\angle IBO&lt;/math&gt;, &lt;math&gt;IH=IO&lt;/math&gt; by Inscribed Angles.<br /> <br /> There are two ways to proceed. <br /> <br /> <br /> Letting &lt;math&gt;O'&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; be the circumcenter and circumradius, respectively, of cyclic pentagon &lt;math&gt;BCOIH&lt;/math&gt;, the most straightforward is to write &lt;math&gt;[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]&lt;/math&gt;, whence &lt;cmath&gt;[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))&lt;/cmath&gt; and, using the fact that &lt;math&gt;R&lt;/math&gt; is fixed, maximize &lt;math&gt;2\sin(60-C)+\sin(2C-60)&lt;/math&gt; with Jensen's Inequality. <br /> <br /> <br /> A more elegant way is shown below.<br /> <br /> '''Lemma:''' &lt;math&gt;[BOIH]&lt;/math&gt; is maximized only if &lt;math&gt;HB=HI&lt;/math&gt;.<br /> <br /> '''Proof by contradiction:''' Suppose &lt;math&gt;[BOIH]&lt;/math&gt; is maximized when &lt;math&gt;HB\neq HI&lt;/math&gt;. Let &lt;math&gt;H'&lt;/math&gt; be the midpoint of minor arc &lt;math&gt;BI&lt;/math&gt; be and &lt;math&gt;I'&lt;/math&gt; the midpoint of minor arc &lt;math&gt;H'O&lt;/math&gt;. Then &lt;math&gt;[BOIH']=[IBO]+[IBH']&gt;[IBO]+[IBH]=[BOIH]&lt;/math&gt; since the altitude from &lt;math&gt;H'&lt;/math&gt; to &lt;math&gt;BI&lt;/math&gt; is greater than that from &lt;math&gt;H&lt;/math&gt; to &lt;math&gt;BI&lt;/math&gt;; similarly &lt;math&gt;[BH'I'O]&gt;[BOIH']&gt;[BOIH]&lt;/math&gt;. Taking &lt;math&gt;H'&lt;/math&gt;, &lt;math&gt;I'&lt;/math&gt; to be the new orthocenter, incenter, respectively, this contradicts the maximality of &lt;math&gt;[BOIH]&lt;/math&gt;, so our claim follows. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> With our lemma(&lt;math&gt;HB=HI&lt;/math&gt;) and &lt;math&gt;IH=IO&lt;/math&gt; from above, along with the fact that inscribed angles that intersect the same length chords are equal,<br /> &lt;cmath&gt;\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}&lt;/cmath&gt; <br /> <br /> -Solution by '''thecmd999'''<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}}<br /> {{MAA Notice}}</div> Owjebra https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_25&diff=128591 2011 AMC 12A Problems/Problem 25 2020-07-18T19:45:04Z <p>Owjebra: /* Solution */</p> <hr /> <div>== Problem ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;\angle BAC = 60^{\circ}&lt;/math&gt;, &lt;math&gt;\angle CBA \leq 90^{\circ}&lt;/math&gt;, &lt;math&gt;BC=1&lt;/math&gt;, and &lt;math&gt;AC \geq AB&lt;/math&gt;. Let &lt;math&gt;H&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, and &lt;math&gt;O&lt;/math&gt; be the orthocenter, incenter, and circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, respectively. Assume that the area of pentagon &lt;math&gt;BCOIH&lt;/math&gt; is the maximum possible. What is &lt;math&gt;\angle CBA&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 60^{\circ} \qquad<br /> \textbf{(B)}\ 72^{\circ} \qquad<br /> \textbf{(C)}\ 75^{\circ} \qquad<br /> \textbf{(D)}\ 80^{\circ} \qquad<br /> \textbf{(E)}\ 90^{\circ} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;\angle CAB=A&lt;/math&gt;, &lt;math&gt;\angle ABC=B&lt;/math&gt;, &lt;math&gt;\angle BCA=C&lt;/math&gt; for convenience.<br /> <br /> It's well-known that &lt;math&gt;\angle BOC=2A&lt;/math&gt;, &lt;math&gt;\angle BIC=90+\frac{A}{2}&lt;/math&gt;, and &lt;math&gt;\angle BHC=180-A&lt;/math&gt; (verifiable by angle chasing). Then, as &lt;math&gt;A=60&lt;/math&gt;, it follows that &lt;math&gt;\angle BOC=\angle BIC=\angle BHC=120&lt;/math&gt; and consequently pentagon &lt;math&gt;BCOIH&lt;/math&gt; is cyclic. Observe that &lt;math&gt;BC=1&lt;/math&gt; is fixed, whence the circumcircle of cyclic pentagon &lt;math&gt;BCOIH&lt;/math&gt; is also fixed. Similarly, as &lt;math&gt;OB=OC&lt;/math&gt;(both are radii), it follows that &lt;math&gt;O&lt;/math&gt; and also &lt;math&gt;[BCO]&lt;/math&gt; is fixed. Since &lt;math&gt;[BCOIH]=[BCO]+[BOIH]&lt;/math&gt; is maximal, it suffices to maximize &lt;math&gt;[BOIH]&lt;/math&gt;.<br /> <br /> Verify that &lt;math&gt;\angle IBC=\frac{B}{2}&lt;/math&gt;, &lt;math&gt;\angle HBC=90-C&lt;/math&gt; by angle chasing; it follows that &lt;math&gt;\angle IBH=\angle HBC-\angle IBC=90-C-\frac{B}{2}=\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}&lt;/math&gt; since &lt;math&gt;A+B+C=180\implies\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90&lt;/math&gt; by Triangle Angle Sum. Similarly, &lt;math&gt;\angle OBC=(180-120)/2=30&lt;/math&gt; (isosceles base angles are equal), hence &lt;cmath&gt;\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}&lt;/cmath&gt; <br /> Since &lt;math&gt;\angle IBH=\angle IBO&lt;/math&gt;, &lt;math&gt;IH=IO&lt;/math&gt; by Inscribed Angles.<br /> <br /> There are two ways to proceed. <br /> <br /> <br /> Letting &lt;math&gt;O'&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; be the circumcenter and circumradius, respectively, of cyclic pentagon &lt;math&gt;BCOIH&lt;/math&gt;, the most straightforward is to write &lt;math&gt;[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]&lt;/math&gt;, whence &lt;cmath&gt;[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))&lt;/cmath&gt; and, using the fact that &lt;math&gt;R&lt;/math&gt; is fixed, maximize &lt;math&gt;2\sin(60-C)+\sin(2C-60)&lt;/math&gt; with Jensen's Inequality. <br /> <br /> <br /> A more elegant way is shown below.<br /> <br /> '''Lemma:''' &lt;math&gt;[BOIH]&lt;/math&gt; is maximized only if &lt;math&gt;HB=HI&lt;/math&gt;.<br /> <br /> '''Proof by contradiction:''' Suppose &lt;math&gt;[BOIH]&lt;/math&gt; is maximized when &lt;math&gt;HB\neq HI&lt;/math&gt;. Let &lt;math&gt;H'&lt;/math&gt; be the midpoint of minor arc &lt;math&gt;BI&lt;/math&gt; be and &lt;math&gt;I'&lt;/math&gt; the midpoint of minor arc &lt;math&gt;H'O&lt;/math&gt;. Then &lt;math&gt;[BOIH']=[IBO]+[IBH']&gt;[IBO]+[IBH]=[BOIH]&lt;/math&gt; since the altitude from &lt;math&gt;H'&lt;/math&gt; to &lt;math&gt;BI&lt;/math&gt; is greater than that from &lt;math&gt;H&lt;/math&gt; to &lt;math&gt;BI&lt;/math&gt;; similarly &lt;math&gt;[BH'I'O]&gt;[BOIH']&gt;[BOIH]&lt;/math&gt;. Taking &lt;math&gt;H'&lt;/math&gt;, &lt;math&gt;I'&lt;/math&gt; to be the new orthocenter, incenter, respectively, this contradicts the maximality of &lt;math&gt;[BOIH]&lt;/math&gt;, so our claim follows. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> With our lemma(&lt;math&gt;HB=HI&lt;/math&gt;) and &lt;math&gt;IH=IO&lt;/math&gt; from above, along with the fact that inscribed angles that intersect the same length chords are equal,<br /> &lt;cmath&gt;\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{(D)}&lt;/cmath&gt; <br /> <br /> -Solution by '''thecmd999'''<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}}<br /> {{MAA Notice}}</div> Owjebra https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:Competition_ratings&diff=128471 AoPS Wiki:Competition ratings 2020-07-17T17:35:24Z <p>Owjebra: /* HMMT (November) */ The first five problems are relatively easy.</p> <hr /> <div>This page contains an approximate estimation of the difficulty level of various [[List of mathematics competitions|competitions]]. It is designed with the intention of introducing contests of similar difficulty levels (but possibly different styles of problems) that readers may like to try to gain more experience.<br /> <br /> Each entry groups the problems into sets of similar difficulty levels and suggests an approximate difficulty rating, on a scale from 1 to 10 (from easiest to hardest). Note that many of these ratings are not directly comparable, because the actual competitions have many different rules; the ratings are generally synchronized with the amount of available time, etc. Also, due to variances within a contest, ranges shown may overlap. A sample problem is provided with each entry, with a link to a solution. <br /> <br /> As you may have guessed with time many competitions got more challenging because many countries got more access to books targeted at olympiad preparation. But especially web site where one can discuss Olympiads such as our very own AoPS!<br /> <br /> If you have some experience with mathematical competitions, we hope that you can help us make the difficulty rankings more accurate. Currently, the system is on a scale from 1 to 10 where 1 is the easiest level, e.g. [http://www.mathlinks.ro/resources.php?c=182&amp;cid=44 early AMC problems] and 10 is hardest level, e.g. [http://www.mathlinks.ro/resources.php?c=37&amp;cid=47 China IMO Team Selection Test.] When considering problem difficulty '''put more emphasis on problem-solving aspects and less so on technical skill requirements'''.<br /> <br /> = Scale =<br /> All levels are estimated and refer to ''averages''. The following is a rough standard based on the USA tier system AMC 8 – AMC 10 – AMC 12 – AIME – USAMO/USAJMO, representing Middle School – Junior High – High School – Challenging High School – Olympiad levels. Other contests can be interpolated against this. <br /> # Problems strictly for beginner, on the easiest elementary school or middle school levels (MOEMS, easy Mathcounts questions, #1-20 on AMC 8s, #1-10 AMC 10s, and others that involve standard techniques introduced up to the middle school level), most traditional middle/high school word problems<br /> # For motivated beginners, harder questions from the previous categories (#21-25 on AMC 8, Challenging Mathcounts questions, #11-20 on AMC 10, #5-10 on AMC 12, the easiest AIME questions, etc), traditional middle/high school word problems with extremely complex problem solving<br /> # Beginner/novice problems that require more creative thinking (MathCounts National, #21-25 on AMC 10, #11-20ish on AMC 12, easier #1-5 on AIMEs, etc.)<br /> # Intermediate-leveled problems, the most difficult questions on AMC 12s (#21-25s), more difficult AIME-styled questions such as #6-9.<br /> # More difficult AIME problems (#10-12), simple proof-based problems (JBMO), etc<br /> # High-leveled AIME-styled questions (#13-15). Introductory-leveled Olympiad-level questions (#1,4s).<br /> # Tougher Olympiad-level questions, #1,4s that require more technical knowledge than new students to Olympiad-type questions have, easier #2,5s, etc.<br /> # High-level Olympiad-level questions, eg #2,5s on difficult Olympiad contest and easier #3,6s, etc.<br /> # Expert Olympiad-level questions, eg #3,6s on difficult Olympiad contests.<br /> # Super Expert problems, problems occasionally even unsuitable for very hard competitions (like the IMO) due to being exceedingly tedious/long/difficult (e.g. very few students are capable of solving, even on a worldwide basis).<br /> <br /> = Competitions =<br /> <br /> ==Introductory Competitions==<br /> Most middle school and first-stage high school competitions would fall under this category. Problems in these competitions are usually ranked from 1 to 3. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AIntroductory+mathematics+competitions here].<br /> <br /> === [[MOEMS]] ===<br /> *Division E: '''1'''<br /> *: ''The whole number &lt;math&gt;N&lt;/math&gt; is divisible by &lt;math&gt;7&lt;/math&gt;. &lt;math&gt;N&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;2,3,4,&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. What is the smallest value that &lt;math&gt;N&lt;/math&gt; can be?'' ([http://www.moems.org/sample_files/SampleE.pdf Solution])<br /> *Division M: '''1'''<br /> *: ''The value of a two-digit number is &lt;math&gt;10&lt;/math&gt; times more than the sum of its digits. The units digit is 1 more than twice the tens digit. Find the two-digit number.'' ([http://www.moems.org/sample_files/SampleM.pdf Solution])<br /> <br /> === [[AMC 8]] ===<br /> <br /> * Problem 1 - Problem 12: '''1''' <br /> *: ''The &lt;math&gt;\emph{harmonic mean}&lt;/math&gt; of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?'' ([[2018 AMC 8 Problems/Problem 10|Solution]])<br /> * Problem 13 - Problem 25: '''1.5-2'''<br /> *: ''How many positive factors does &lt;math&gt;23,232&lt;/math&gt; have?'' ([[2018 AMC 8 Problems/Problem 18|Solution]])<br /> <br /> === [[Mathcounts]] ===<br /> <br /> * Countdown: '''1-2.'''<br /> * Sprint: '''1-1.5''' (school/chapter), '''1.5-2''' (State), '''2-2.5''' (National)<br /> * Target: '''1-2''' (school/chapter), '''1.5-2.5''' (State), '''2-3''' (National)<br /> <br /> === [[AMC 10]] ===<br /> <br /> * Problem 1 - 10: '''1-2'''<br /> *: ''A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?'' ([[2016 AMC 10A Problems/Problem 5|Solution]])<br /> * Problem 11 - 20: '''2-3'''<br /> *: ''For some positive integer &lt;math&gt;k&lt;/math&gt;, the repeating base-&lt;math&gt;k&lt;/math&gt; representation of the (base-ten) fraction &lt;math&gt;\frac{7}{51}&lt;/math&gt; is &lt;math&gt;0.\overline{23}_k = 0.232323..._k&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?'' ([[2019 AMC 10A Problems/Problem 18|Solution]])<br /> * Problem 21 - 25: '''3.5-4.5'''<br /> *: ''The vertices of an equilateral triangle lie on the hyperbola &lt;math&gt;xy=1&lt;/math&gt;, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?'' ([[2017 AMC 10B Problems/Problem 24|Solution]])<br /> <br /> ===[[CEMC|CEMC Multiple Choice Tests]]===<br /> This covers the CEMC Gauss, Pascal, Cayley, and Fermat tests.<br /> <br /> * Part A: '''0.5-1.5'''<br /> *: ''How many different 3-digit whole numbers can be formed using the digits 4, 7, and 9, assuming that no digit can be repeated in a number?'' (2015 Gauss 7 Problem 10)<br /> * Part B: '''1-2'''<br /> *: ''Two lines with slopes &lt;math&gt;\tfrac14&lt;/math&gt; and &lt;math&gt;\tfrac54&lt;/math&gt; intersect at &lt;math&gt;(1,1)&lt;/math&gt;. What is the area of the triangle formed by these two lines and the vertical line &lt;math&gt;x = 5&lt;/math&gt;?'' (2017 Cayley Problem 19)<br /> * Part C (Gauss/Pascal): '''2-2.5'''<br /> *: ''Suppose that &lt;math&gt;\tfrac{2009}{2014} + \tfrac{2019}{n} = \tfrac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;n&lt;/math&gt; are positive integers with &lt;math&gt;\tfrac{a}{b}&lt;/math&gt; in lowest terms. What is the sum of the digits of the smallest positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;a&lt;/math&gt; is a multiple of 1004?'' (2014 Pascal Problem 25)<br /> * Part C (Cayley/Fermat): '''2.5-3'''<br /> *: ''Wayne has 3 green buckets, 3 red buckets, 3 blue buckets, and 3 yellow buckets. He randomly distributes 4 hockey pucks among the green buckets, with each puck equally likely to be put in each bucket. Similarly, he distributes 3 pucks among the red buckets, 2 pucks among the blue buckets, and 1 puck among the yellow buckets. Once he is ﬁnished, what is the probability that a green bucket contains more pucks than each of the other 11 buckets?'' (2018 Fermat Problem 24)<br /> <br /> ===[[CEMC|CEMC Fryer/Galois/Hypatia]]===<br /> <br /> * Problem 1-2: '''1-2'''<br /> * Problem 3-4 (early parts): '''2-3'''<br /> * Problem 3-4 (later parts): '''3-5'''<br /> <br /> ===Problem Solving Books for Introductory Students===<br /> <br /> Remark: There are many other problem books for Introductory Students that are not published by AoPS. Typically the rating on the left side is equivalent to the difficulty of the easiest review problems and the difficulty on the right side is the difficulty of the hardest challenge problems. The difficulty may vary greatly between sections of a book.<br /> <br /> ===[[Prealgebra by AoPS]]===<br /> 1-2<br /> ===[[Introduction to Algebra by AoPS]]===<br /> 1-3.5<br /> ===[[Introduction to Counting and Probability by AoPS]]===<br /> 1-3.5<br /> ===[[Introduction to Number Theory by AoPS]]===<br /> 1-3<br /> ===[[Introduction to Geometry by AoPS]]===<br /> 1-4<br /> <br /> ==Intermediate Competitions==<br /> This category consists of all the non-proof math competitions for the middle stages of high school. The difficulty range would normally be from 3 to 6. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AIntermediate+mathematics+competitions here].<br /> <br /> === [[AMC 12]] ===<br /> <br /> * Problem 1-10: '''1.5-2'''<br /> *: ''What is the value of &lt;cmath&gt;\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?&lt;/cmath&gt;'' ([[2018 AMC 12B Problems/Problem 7|Solution]])<br /> * Problem 11-20: '''2.5-3.5'''<br /> *: ''Points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; lie on a circle centered at &lt;math&gt;O&lt;/math&gt;, and &lt;math&gt;\angle AOB = 60^\circ&lt;/math&gt;. A second circle is internally tangent to the first and tangent to both &lt;math&gt;\overline{OA}&lt;/math&gt; and &lt;math&gt;\overline{OB}&lt;/math&gt;. What is the ratio of the area of the smaller circle to that of the larger circle?'' ([[2008 AMC 12A Problems/Problem 13|Solution]])<br /> * Problem 21-25: '''4-5.5'''<br /> *: ''Functions &lt;math&gt;f&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; are quadratic, &lt;math&gt;g(x) = - f(100 - x)&lt;/math&gt;, and the graph of &lt;math&gt;g&lt;/math&gt; contains the vertex of the graph of &lt;math&gt;f&lt;/math&gt;. The four &lt;math&gt;x&lt;/math&gt;-intercepts on the two graphs have &lt;math&gt;x&lt;/math&gt;-coordinates &lt;math&gt;x_1&lt;/math&gt;, &lt;math&gt;x_2&lt;/math&gt;, &lt;math&gt;x_3&lt;/math&gt;, and &lt;math&gt;x_4&lt;/math&gt;, in increasing order, and &lt;math&gt;x_3 - x_2 = 150&lt;/math&gt;. The value of &lt;math&gt;x_4 - x_1&lt;/math&gt; is &lt;math&gt;m + n\sqrt p&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. What is &lt;math&gt;m + n + p&lt;/math&gt;?'' ([[2009 AMC 12A Problems/Problem 23|Solution]])<br /> <br /> === [[AIME]] ===<br /> <br /> * Problem 1 - 5: '''3-3.5'''<br /> *: ''Consider the integer &lt;cmath&gt;N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.&lt;/cmath&gt;Find the sum of the digits of &lt;math&gt;N&lt;/math&gt;.'' ([[2019 AIME I Problems/Problem 1|Solution]])<br /> * Problem 6 - 9: '''4-4.5''' <br /> *: ''Define &lt;math&gt;n!!&lt;/math&gt; to be &lt;math&gt;n(n-2)(n-4)\cdots 3\cdot 1&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt; odd and &lt;math&gt;n(n-2)(n-4)\cdots 4\cdot 2&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt; even. When &lt;math&gt;\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}&lt;/math&gt; is expressed as a fraction in lowest terms, its denominator is &lt;math&gt;2^ab&lt;/math&gt; with &lt;math&gt;b&lt;/math&gt; odd. Find &lt;math&gt;\dfrac{ab}{10}&lt;/math&gt;.'' ([[2009 AIME II Problems/Problem 7|Solution]])<br /> * Problem 10 - 12: '''5-5.5'''<br /> *: Let &lt;math&gt;R&lt;/math&gt; be the set of all possible remainders when a number &lt;math&gt;2^n&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; a nonnegative integer, is divided by &lt;math&gt;1000&lt;/math&gt;.Let &lt;math&gt;S&lt;/math&gt; be the sum of all elements in &lt;math&gt;R&lt;/math&gt;. Find the remainder when &lt;math&gt;S&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt; ([[2011 AIME I Problems/Problem 11|Solution]])<br /> * Problem 13 - 15: '''6-6.5'''<br /> *: ''Let<br /> <br /> &lt;cmath&gt;P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).&lt;/cmath&gt;<br /> Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;i = \sqrt { - 1},&lt;/math&gt; and &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> <br /> &lt;cmath&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/cmath&gt;<br /> where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;.'' ([[2003 AIME II Problems/Problem 15|Solution]])<br /> <br /> === [[ARML]] ===<br /> <br /> * Individuals, Problem 1: '''2'''<br /> <br /> * Individuals, Problems 2, 3, 4, 5, 7, and 9: '''3'''<br /> <br /> * Individuals, Problems 6 and 8: '''4''' <br /> <br /> * Individuals, Problem 10: '''5.5'''<br /> <br /> * Team/power, Problem 1-5: '''3.5''' <br /> <br /> * Team/power, Problem 6-10: '''5'''<br /> <br /> ===[[HMMT|HMMT (November)]]===<br /> * Individual Round, Problem 6-8: '''4'''<br /> * Individual Round, Problem 10: '''4.5'''<br /> * Team Round: '''4-5'''<br /> * Guts: '''3.5-5.25'''<br /> <br /> ===[[CEMC|CEMC Euclid]]===<br /> <br /> * Problem 1-6: '''1-3'''<br /> * Problem 7-10: '''3-5'''<br /> <br /> ===[[Purple Comet! Math Meet|Purple Comet]]===<br /> <br /> * Problems 1-10 (MS): '''1.5-2.5'''<br /> * Problems 11-20 (MS): '''2.5-4'''<br /> * Problems 1-10 (HS): '''1.5-2.5'''<br /> * Problems 11-20 (HS): '''2-3.5'''<br /> * Problems 21-30 (HS): '''3.5-4.5'''<br /> <br /> === [[Philippine Mathematical Olympiad Qualifying Round]] ===<br /> <br /> * Problem 1-15: '''2'''<br /> * Problem 16-25: '''3'''<br /> * Problem 26-30: '''4'''<br /> <br /> ==Beginner Olympiad Competitions==<br /> This category consists of beginning Olympiad math competitions. Most junior and first stage Olympiads fall under this category. The range from the difficulty scale would be around 4 to 6. A full list is available [http://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3ABeginner+Olympiad+mathematics+competitions here].<br /> <br /> === [[USAMTS]] ===<br /> USAMTS generally has a different feel to it than olympiads, and is mainly for proofwriting practice instead of olympiad practice depending on how one takes the test. USAMTS allows an entire month to solve problems, with internet resources and books being allowed. However, the ultimate gap is that it permits computer programs to be used, and that Problem 1 is not a proof problem. However, it can still be roughly put to this rating scale:<br /> * Problem 1-2: '''3-4'''<br /> *: ''Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter.'' ([http://usamts.org/Solutions/Solution2_3_16.pdf Solution])<br /> * Problem 3-5: '''4-6'''<br /> *: ''Call a positive real number groovy if it can be written in the form &lt;math&gt;\sqrt{n} + \sqrt{n + 1}&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;. Show that if &lt;math&gt;x&lt;/math&gt; is groovy, then for any positive integer &lt;math&gt;r&lt;/math&gt;, the number &lt;math&gt;x^r&lt;/math&gt; is groovy as well.'' ([http://usamts.org/Solutions/Solutions_20_1.pdf Solution])<br /> <br /> === [[Indonesia Mathematical Olympiad|Indonesia MO]] ===<br /> * Problem 1/5: '''3.5'''<br /> *: '' In a drawer, there are at most &lt;math&gt;2009&lt;/math&gt; balls, some of them are white, the rest are blue, which are randomly distributed. If two balls were taken at the same time, then the probability that the balls are both blue or both white is &lt;math&gt;\frac12&lt;/math&gt;. Determine the maximum amount of white balls in the drawer, such that the probability statement is true?'' &lt;url&gt;viewtopic.php?t=294065 (Solution)&lt;/url&gt;<br /> * Problem 2/6: '''4.5'''<br /> *: ''Find the lowest possible values from the function<br /> &lt;math&gt;f(x) = x^{2008} - 2x^{2007} + 3x^{2006} - 4x^{2005} + 5x^{2004} - \cdots - 2006x^3 + 2007x^2 - 2008x + 2009&lt;/math&gt;<br /> <br /> for any real numbers &lt;math&gt;x&lt;/math&gt;.''&lt;url&gt;viewtopic.php?t=294067 (Solution)&lt;/url&gt;<br /> * Problem 3/7: '''5'''<br /> *: ''A pair of integers &lt;math&gt;(m,n)&lt;/math&gt; is called ''good'' if<br /> &lt;math&gt;m\mid n^2 + n \ \text{and} \ n\mid m^2 + m&lt;/math&gt;<br /> <br /> Given 2 positive integers &lt;math&gt;a,b &gt; 1&lt;/math&gt; which are relatively prime, prove that there exists a ''good'' pair &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;a\mid m&lt;/math&gt; and &lt;math&gt;b\mid n&lt;/math&gt;, but &lt;math&gt;a\nmid n&lt;/math&gt; and &lt;math&gt;b\nmid m&lt;/math&gt;.'' &lt;url&gt;viewtopic.php?t=294068 (Solution)&lt;/url&gt;<br /> * Problem 4/8: '''6'''<br /> *: ''Given an acute triangle &lt;math&gt;ABC&lt;/math&gt;. The incircle of triangle &lt;math&gt;ABC&lt;/math&gt; touches &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively at &lt;math&gt;D,E,F&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle A&lt;/math&gt; cuts &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;DF&lt;/math&gt; respectively at &lt;math&gt;K&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt;. Suppose &lt;math&gt;AA_1&lt;/math&gt; is one of the altitudes of triangle &lt;math&gt;ABC&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> (a) Prove that &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CL&lt;/math&gt; are perpendicular with the angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt;.<br /> <br /> (b) Show that &lt;math&gt;A_1KML&lt;/math&gt; is a cyclic quadrilateral.'' &lt;url&gt;viewtopic.php?t=294069 (Solution)&lt;/url&gt;<br /> <br /> === [[Central American Olympiad]] ===<br /> * Problem 1: '''4'''<br /> *: ''Find all three-digit numbers &lt;math&gt;abc&lt;/math&gt; (with &lt;math&gt;a \neq 0&lt;/math&gt;) such that &lt;math&gt;a^{2} + b^{2} + c^{2}&lt;/math&gt; is a divisor of 26.'' (&lt;url&gt;viewtopic.php?p=903856#903856 Solution&lt;/url&gt;)<br /> * Problem 2,4,5: '''5-6'''<br /> *: ''Show that the equation &lt;math&gt;a^{2}b^{2} + b^{2}c^{2} + 3b^{2} - c^{2} - a^{2} = 2005&lt;/math&gt; has no integer solutions.'' (&lt;url&gt;viewtopic.php?p=291301#291301 Solution&lt;/url&gt;)<br /> * Problem 3/6: '''6.5''' <br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral. &lt;math&gt;I = AC\cap BD&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are points on &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CD&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt; respectively, such that &lt;math&gt;EF \cap GH = I&lt;/math&gt;. If &lt;math&gt;M = EG \cap AC&lt;/math&gt;, &lt;math&gt;N = HF \cap AC&lt;/math&gt;, show that &lt;math&gt;\frac {AM}{IM}\cdot \frac {IN}{CN} = \frac {IA}{IC}&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=828841#p828841 Solution&lt;/url&gt;<br /> <br /> === [[JBMO]] ===<br /> <br /> * Problem 1: '''4'''<br /> *: ''Find all real numbers &lt;math&gt;a,b,c,d&lt;/math&gt; such that <br /> &lt;cmath&gt; \left\{\begin{array}{cc}a+b+c+d = 20,\\ ab+ac+ad+bc+bd+cd = 150.\end{array}\right. &lt;/cmath&gt;''<br /> * Problem 2: '''4.5-5'''<br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral with &lt;math&gt;\angle DAC=\angle BDC=36^\circ&lt;/math&gt;, &lt;math&gt;\angle CBD=18^\circ&lt;/math&gt; and &lt;math&gt;\angle BAC=72^\circ&lt;/math&gt;. The diagonals intersect at point &lt;math&gt;P&lt;/math&gt;. Determine the measure of &lt;math&gt;\angle APD&lt;/math&gt;.''<br /> * Problem 3: '''5'''<br /> *: ''Find all prime numbers &lt;math&gt;p,q,r&lt;/math&gt;, such that &lt;math&gt;\frac pq-\frac4{r+1}=1&lt;/math&gt;.''<br /> * Problem 4: '''6'''<br /> *: ''A &lt;math&gt;4\times4&lt;/math&gt; table is divided into &lt;math&gt;16&lt;/math&gt; white unit square cells. Two cells are called neighbors if they share a common side. A '''move''' consists in choosing a cell and changing the colors of neighbors from white to black or from black to white. After exactly &lt;math&gt;n&lt;/math&gt; moves all the &lt;math&gt;16&lt;/math&gt; cells were black. Find all possible values of &lt;math&gt;n&lt;/math&gt;.''<br /> <br /> ==Olympiad Competitions==<br /> This category consists of standard Olympiad competitions, usually ones from national Olympiads. Average difficulty is from 5 to 8. A full list is available [http://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AOlympiad+mathematics+competitions here].<br /> <br /> === [[USAJMO]] ===<br /> * Problem 1/4: '''5'''<br /> *: ''There are &lt;math&gt;a+b&lt;/math&gt; bowls arranged in a row, numbered &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;a+b&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are given positive integers. Initially, each of the first &lt;math&gt;a&lt;/math&gt; bowls contains an apple, and each of the last &lt;math&gt;b&lt;/math&gt; bowls contains a pear.''<br /> <br /> ''A legal move consists of moving an apple from bowl &lt;math&gt;i&lt;/math&gt; to bowl &lt;math&gt;i+1&lt;/math&gt; and a pear from bowl &lt;math&gt;j&lt;/math&gt; to bowl &lt;math&gt;j-1&lt;/math&gt;, provided that the difference &lt;math&gt;i-j&lt;/math&gt; is even. We permit multiple fruits in the same bowl at the same time. The goal is to end up with the first &lt;math&gt;b&lt;/math&gt; bowls each containing a pear and the last &lt;math&gt;a&lt;/math&gt; bowls each containing an apple. Show that this is possible if and only if the product &lt;math&gt;ab&lt;/math&gt; is even.'' ([[2019 USAJMO Problems/Problem 1|Solution]])<br /> <br /> * Problem 2/5: '''6'''<br /> *: ''Let &lt;math&gt;a,b,c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a+b+c=4\sqrt{abc}&lt;/math&gt;. Prove that &lt;cmath&gt;2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.&lt;/cmath&gt;'' ([[2018 USAJMO Problems/Problem 2|Solution]])<br /> <br /> * Problem 3/6: '''7'''<br /> *: ''Two rational numbers &lt;math&gt;\tfrac{m}{n}&lt;/math&gt; and &lt;math&gt;\tfrac{n}{m}&lt;/math&gt; are written on a blackboard, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. At any point, Evan may pick two of the numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; written on the board and write either their arithmetic mean &lt;math&gt;\tfrac{x+y}{2}&lt;/math&gt; or their harmonic mean &lt;math&gt;\tfrac{2xy}{x+y}&lt;/math&gt; on the board as well. Find all pairs &lt;math&gt;(m,n)&lt;/math&gt; such that Evan can write &lt;math&gt;1&lt;/math&gt; on the board in finitely many steps.'' ([[2019 USAJMO Problems/Problem 6|Solution]])<br /> <br /> ===[[HMMT|HMMT (February)]]===<br /> * Individual Round, Problem 1-5: '''5'''<br /> * Individual Round, Problem 6-10: '''5.5-6'''<br /> * Team Round: '''7.5'''<br /> * HMIC: '''8'''<br /> <br /> === [[Canadian MO]] ===<br /> <br /> * Problem 1: '''5.5'''<br /> * Problem 2: '''6'''<br /> * Problem 3: '''6.5''' <br /> * Problem 4: '''7-7.5'''<br /> * Problem 5: '''7.5-8'''<br /> <br /> === Austrian MO ===<br /> <br /> * Regional Competition for Advanced Students, Problems 1-4: '''5''' <br /> * Federal Competition for Advanced Students, Part 1. Problems 1-4: '''6''' <br /> * Federal Competition for Advanced Students, Part 2, Problems 1-6: '''7'''<br /> <br /> === [[Ibero American Olympiad]] ===<br /> <br /> * Problem 1/4: '''5.5'''<br /> * Problem 2/5: '''6.5'''<br /> * Problem 3/6: '''7.5'''<br /> <br /> === [[APMO]] ===<br /> *Problem 1: '''6'''<br /> *Problem 2: '''7'''<br /> *Problem 3: '''7'''<br /> *Problem 4: '''7.5'''<br /> *Problem 5: '''8.5'''<br /> <br /> === Balkan MO ===<br /> <br /> * Problem 1: '''6'''<br /> *: '' Solve the equation &lt;math&gt;3^x - 5^y = z^2&lt;/math&gt; in positive integers. '' <br /> * Problem 2: '''6.5'''<br /> *: '' Let &lt;math&gt;MN&lt;/math&gt; be a line parallel to the side &lt;math&gt;BC&lt;/math&gt; of a triangle &lt;math&gt;ABC&lt;/math&gt;, with &lt;math&gt;M&lt;/math&gt; on the side &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; on the side &lt;math&gt;AC&lt;/math&gt;. The lines &lt;math&gt;BN&lt;/math&gt; and &lt;math&gt;CM&lt;/math&gt; meet at point &lt;math&gt;P&lt;/math&gt;. The circumcircles of triangles &lt;math&gt;BMP&lt;/math&gt; and &lt;math&gt;CNP&lt;/math&gt; meet at two distinct points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Prove that &lt;math&gt;\angle BAQ = \angle CAP&lt;/math&gt;. ''<br /> * Problem 3: '''7.5'''<br /> *: '' A &lt;math&gt;9 \times 12&lt;/math&gt; rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres &lt;math&gt;C_1,C_2...,C_{96}&lt;/math&gt; in such way that the following to conditions are both fulfilled<br /> <br /> &lt;math&gt;(i)&lt;/math&gt; the distances &lt;math&gt;C_1C_2,...C_{95}C_{96}, C_{96}C_{1}&lt;/math&gt; are all equal to &lt;math&gt;\sqrt {13}&lt;/math&gt;<br /> <br /> &lt;math&gt;(ii)&lt;/math&gt; the closed broken line &lt;math&gt;C_1C_2...C_{96}C_1&lt;/math&gt; has a centre of symmetry? ''<br /> * Problem 4: '''8'''<br /> *: '' Denote by &lt;math&gt;S&lt;/math&gt; the set of all positive integers. Find all functions &lt;math&gt;f: S \rightarrow S&lt;/math&gt; such that<br /> <br /> &lt;math&gt;f \bigg(f^2(m) + 2f^2(n)\bigg) = m^2 + 2 n^2&lt;/math&gt; for all &lt;math&gt;m,n \in S&lt;/math&gt;. '<br /> <br /> ==Hard Olympiad Competitions==<br /> This category consists of harder Olympiad contests. Difficulty is usually from 7 to 10. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AHard+Olympiad+mathematics+competitions here].<br /> <br /> === [[USAMO]] ===<br /> * Problem 1/4: '''6-7'''<br /> *: ''Let &lt;math&gt;\mathcal{P}&lt;/math&gt; be a convex polygon with &lt;math&gt;n&lt;/math&gt; sides, &lt;math&gt;n\ge3&lt;/math&gt;. Any set of &lt;math&gt;n - 3&lt;/math&gt; diagonals of &lt;math&gt;\mathcal{P}&lt;/math&gt; that do not intersect in the interior of the polygon determine a ''triangulation'' of &lt;math&gt;\mathcal{P}&lt;/math&gt; into &lt;math&gt;n - 2&lt;/math&gt; triangles. If &lt;math&gt;\mathcal{P}&lt;/math&gt; is regular and there is a triangulation of &lt;math&gt;\mathcal{P}&lt;/math&gt; consisting of only isosceles triangles, find all the possible values of &lt;math&gt;n&lt;/math&gt;.'' ([[2008 USAMO Problems/Problem 4|Solution]]) <br /> * Problem 2/5: '''7-8'''<br /> *: ''Three nonnegative real numbers &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, &lt;math&gt;r_3&lt;/math&gt; are written on a blackboard. These numbers have the property that there exist integers &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, not all zero, satisfying &lt;math&gt;a_1r_1 + a_2r_2 + a_3r_3 = 0&lt;/math&gt;. We are permitted to perform the following operation: find two numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; on the blackboard with &lt;math&gt;x \le y&lt;/math&gt;, then erase &lt;math&gt;y&lt;/math&gt; and write &lt;math&gt;y - x&lt;/math&gt; in its place. Prove that after a finite number of such operations, we can end up with at least one &lt;math&gt;0&lt;/math&gt; on the blackboard.'' ([[2008 USAMO Problems/Problem 5|Solution]])<br /> * Problem 3/6: '''8-9'''<br /> *: ''Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree &lt;math&gt;n &lt;/math&gt; with real coefficients is the average of two monic polynomials of degree &lt;math&gt;n &lt;/math&gt; with &lt;math&gt;n &lt;/math&gt; real roots.'' ([[2002 USAMO Problems/Problem 3|Solution]])<br /> <br /> === [[USA TST]] ===<br /> <br /> <br /> <br /> * Problem 1/4/7: '''6.5-7'''<br /> * Problem 2/5/8: '''7.5-8'''<br /> * Problem 3/6/9: '''8.5-9'''<br /> <br /> === [[Putnam]] ===<br /> <br /> * Problem A/B,1-2: '''7'''<br /> *: ''Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola &lt;math&gt;xy = 1&lt;/math&gt; and both branches of the hyperbola &lt;math&gt;xy = - 1.&lt;/math&gt; (A set &lt;math&gt;S&lt;/math&gt; in the plane is called ''convex'' if for any two points in &lt;math&gt;S&lt;/math&gt; the line segment connecting them is contained in &lt;math&gt;S.&lt;/math&gt;)'' ([https://artofproblemsolving.com/community/c7h177227p978383 Solution])<br /> * Problem A/B,3-4: '''8'''<br /> *: ''Let &lt;math&gt;H&lt;/math&gt; be an &lt;math&gt;n\times n&lt;/math&gt; matrix all of whose entries are &lt;math&gt;\pm1&lt;/math&gt; and whose rows are mutually orthogonal. Suppose &lt;math&gt;H&lt;/math&gt; has an &lt;math&gt;a\times b&lt;/math&gt; submatrix whose entries are all &lt;math&gt;1.&lt;/math&gt; Show that &lt;math&gt;ab\le n&lt;/math&gt;.'' ([https://artofproblemsolving.com/community/c7h64435p383280 Solution])<br /> * Problem A/B,5-6: '''9'''<br /> *: ''For any &lt;math&gt;a &gt; 0&lt;/math&gt;, define the set &lt;math&gt;S(a) = \{[an]|n = 1,2,3,...\}&lt;/math&gt;. Show that there are no three positive reals &lt;math&gt;a,b,c&lt;/math&gt; such that &lt;math&gt;S(a)\cap S(b) = S(b)\cap S(c) = S(c)\cap S(a) = \emptyset,S(a)\cup S(b)\cup S(c) = \{1,2,3,...\}&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?t=127810 Solution&lt;/url&gt;)<br /> <br /> === [[China TST]] ===<br /> <br /> * Problem 1/4: '''8-8.5''' <br /> *: ''Given an integer &lt;math&gt;m,&lt;/math&gt; prove that there exist odd integers &lt;math&gt;a,b&lt;/math&gt; and a positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;cmath&gt;2m=a^{19}+b^{99}+k*2^{1000}.&lt;/cmath&gt;''<br /> * Problem 2/5: '''9''' <br /> *: ''Given a positive integer &lt;math&gt;n&gt;1&lt;/math&gt; and real numbers &lt;math&gt;a_1 &lt; a_2 &lt; \ldots &lt; a_n,&lt;/math&gt; such that &lt;math&gt;\dfrac{1}{a_1} + \dfrac{1}{a_2} + \ldots + \dfrac{1}{a_n} \le 1,&lt;/math&gt; prove that for any positive real number &lt;math&gt;x,&lt;/math&gt; &lt;cmath&gt;\left(\dfrac{1}{a_1^2+x} + \dfrac{1}{a_2^2+x} + \ldots + \dfrac{1}{a_n^2+x}\right)^2 \ge \dfrac{1}{2a_1(a_1-1)+2x}.&lt;/cmath&gt;''<br /> * Problem 3/6: '''9.5-10'''<br /> *: ''Let &lt;math&gt;n&gt;1&lt;/math&gt; be an integer and let &lt;math&gt;a_0,a_1,\ldots,a_n&lt;/math&gt; be non-negative real numbers. Define &lt;math&gt;S_k=\sum_{i=0}^k \binom{k}{i}a_i&lt;/math&gt; for &lt;math&gt;k=0,1,\ldots,n&lt;/math&gt;. Prove that&lt;cmath&gt;\frac{1}{n} \sum_{k=0}^{n-1} S_k^2-\frac{1}{n^2}\left(\sum_{k=0}^{n} S_k\right)^2\le \frac{4}{45} (S_n-S_0)^2.&lt;/cmath&gt;''<br /> <br /> === [[IMO]] ===<br /> <br /> * Problem 1/4: '''5.5-7'''<br /> *: ''Let &lt;math&gt;\Gamma&lt;/math&gt; be the circumcircle of acute triangle &lt;math&gt;ABC&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are on segments &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; respectively such that &lt;math&gt;AD = AE&lt;/math&gt;. The perpendicular bisectors of &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;CE&lt;/math&gt; intersect minor arcs &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; of &lt;math&gt;\Gamma&lt;/math&gt; at points &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; respectively. Prove that lines &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FG&lt;/math&gt; are either parallel or they are the same line.'' ([[2018 IMO Problems/Problem 1|Solution]])<br /> <br /> * Problem 2/5: '''7-8'''<br /> *: ''Let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial of degree &lt;math&gt;n&gt;1&lt;/math&gt; with integer coefficients, and let &lt;math&gt;k&lt;/math&gt; be a positive integer. Consider the polynomial &lt;math&gt;Q(x) = P( P ( \ldots P(P(x)) \ldots ))&lt;/math&gt;, where &lt;math&gt;P&lt;/math&gt; occurs &lt;math&gt;k&lt;/math&gt; times. Prove that there are at most &lt;math&gt;n&lt;/math&gt; integers &lt;math&gt;t&lt;/math&gt; such that &lt;math&gt;Q(t)=t&lt;/math&gt;.'' ([[2006 IMO Problems/Problem 5|Solution]])<br /> <br /> * Problem 3/6: '''9-10'''<br /> *: ''Assign to each side &lt;math&gt;b&lt;/math&gt; of a convex polygon &lt;math&gt;P&lt;/math&gt; the maximum area of a triangle that has &lt;math&gt;b&lt;/math&gt; as a side and is contained in &lt;math&gt;P&lt;/math&gt;. Show that the sum of the areas assigned to the sides of &lt;math&gt;P&lt;/math&gt; is at least twice the area of &lt;math&gt;P&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=572824#572824 Solution&lt;/url&gt;)<br /> <br /> === [[IMO Shortlist]] ===<br /> <br /> * Problem 1-2: '''5.5-7'''<br /> * Problem 3-4: '''7-8'''<br /> * Problem 5+: '''8-10'''<br /> <br /> [[Category:Mathematics competitions]]</div> Owjebra https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_24&diff=113671 2018 AMC 10A Problems/Problem 24 2019-12-29T06:00:45Z <p>Owjebra: /* Solution 5: Trig */ Changed &quot;area of triangle $ABC$ is equal to $\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}$ to &quot;area of triangle $ADE$...&quot;</p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;AB=50&lt;/math&gt; and &lt;math&gt;AC=10&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;, and let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt; intersects &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;, respectively. What is the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }60 \qquad<br /> \textbf{(B) }65 \qquad<br /> \textbf{(C) }70 \qquad<br /> \textbf{(D) }75 \qquad<br /> \textbf{(E) }80 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Let &lt;math&gt;BC = a&lt;/math&gt;, &lt;math&gt;BG = x&lt;/math&gt;, &lt;math&gt;GC = y&lt;/math&gt;, and the length of the perpendicular to &lt;math&gt;BC&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; be &lt;math&gt;h&lt;/math&gt;. By angle bisector theorem, we have that &lt;cmath&gt;\frac{50}{x} = \frac{10}{y},&lt;/cmath&gt; where &lt;math&gt;y = -x+a&lt;/math&gt;. Therefore substituting we have that &lt;math&gt;BG=\frac{5a}{6}&lt;/math&gt;. By similar triangles, we have that &lt;math&gt;DF=\frac{5a}{12}&lt;/math&gt;, and the height of this trapezoid is &lt;math&gt;\frac{h}{2}&lt;/math&gt;. Then, we have that &lt;math&gt;\frac{ah}{2}=120&lt;/math&gt;. We wish to compute &lt;math&gt;\frac{5a}{8}\cdot\frac{h}{2}&lt;/math&gt;, and we have that it is &lt;math&gt;\boxed{75}&lt;/math&gt; by substituting.<br /> <br /> == Solution 2 ==<br /> <br /> For this problem, we have &lt;math&gt;\triangle{ADE}\sim\triangle{ABC}&lt;/math&gt; because of SAS and &lt;math&gt;DE = \frac{BC}{2}&lt;/math&gt;. Therefore, &lt;math&gt;\bigtriangleup ADE&lt;/math&gt; is a quarter of the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, which is &lt;math&gt;30&lt;/math&gt;. Subsequently, we can compute the area of quadrilateral &lt;math&gt;BDEC&lt;/math&gt; to be &lt;math&gt;120 - 30 = 90&lt;/math&gt;. Using the angle bisector theorem in the same fashion as the previous problem, we get that &lt;math&gt;\overline{BG}&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt; times the length of &lt;math&gt;\overline{GC}&lt;/math&gt;. We want the larger piece, as described by the problem. Because the heights are identical, one area is &lt;math&gt;5&lt;/math&gt; times the other, and &lt;math&gt;\frac{5}{6} \cdot 90 = \boxed{75}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> The area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;5:1&lt;/math&gt; by Law of Sines. So the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt;. Since &lt;math&gt;\overline{DE}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, so &lt;math&gt;\overline{DF}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; . So the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;1:4&lt;/math&gt; , so the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, by similar triangles. Therefore the area of quad &lt;math&gt;FDBG&lt;/math&gt; is &lt;math&gt;100-25=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 4 ==<br /> The area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; is the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; minus the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt;. Notice, &lt;math&gt;\overline{DE} || \overline{BC}&lt;/math&gt;, so &lt;math&gt;\bigtriangleup ABG \sim \bigtriangleup ADF&lt;/math&gt;, and since &lt;math&gt;\overline{AD}:\overline{AB}=1:2&lt;/math&gt;, the area of &lt;math&gt;\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4&lt;/math&gt;. Given that the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt;, using &lt;math&gt;\frac{bh}{2}&lt;/math&gt; on side &lt;math&gt;AB&lt;/math&gt; yields &lt;math&gt;\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}&lt;/math&gt;. Using the Angle Bisector Theorem, &lt;math&gt;\overline{BG}:\overline{BC}=50:(10+50)=5:6&lt;/math&gt;, so the height of &lt;math&gt;\bigtriangleup ABG: \bigtriangleup ACB=5:6&lt;/math&gt;. Therefore our answer is &lt;math&gt;\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 5: Trig ==<br /> We try to find the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; by subtracting the area outside the quadrilateral but inside triangle &lt;math&gt;ABC&lt;/math&gt;. Note that the area of &lt;math&gt;\triangle ADE&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A&lt;/math&gt;. The ratio &lt;math&gt;\frac{[ADE]}{[ABC]}&lt;/math&gt; is thus equal to &lt;math&gt;\frac{1}{4}&lt;/math&gt; and the area of triangle &lt;math&gt;ADE&lt;/math&gt; is &lt;math&gt;\frac{1}{4} \cdot 120 = 30&lt;/math&gt;. Let side &lt;math&gt;BC&lt;/math&gt; be equal to &lt;math&gt;6x&lt;/math&gt;, then &lt;math&gt;BG = 5x, GC = x&lt;/math&gt; by the angle bisector theorem. Similarly, we find the area of triangle &lt;math&gt;AGC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 10 \cdot x \cdot \sin C&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C&lt;/math&gt;. A ratio between these two triangles yields &lt;math&gt;\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}&lt;/math&gt;, so &lt;math&gt;[AGC] = 20&lt;/math&gt;. Now we just need to find the area of triangle &lt;math&gt;AFE&lt;/math&gt; and subtract it from the combined areas of &lt;math&gt;[ADE]&lt;/math&gt; and &lt;math&gt;[ACG]&lt;/math&gt;, since we count it twice. Note that the angle bisector theorem also applies for &lt;math&gt;\triangle ADE&lt;/math&gt; and &lt;math&gt;\frac{AE}{AD} = \frac{1}{5}&lt;/math&gt;, so thus &lt;math&gt;\frac{EF}{ED} = \frac{1}{6}&lt;/math&gt; and we find &lt;math&gt;[AFE] = \frac{1}{6} \cdot 30 = 5&lt;/math&gt;, and the area outside &lt;math&gt;FDBG&lt;/math&gt; must be &lt;math&gt; [ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45&lt;/math&gt;, and we finally find &lt;math&gt;[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}&lt;/math&gt;, and we are done. ~skyscraper<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Owjebra