https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Math101010&feedformat=atom AoPS Wiki - User contributions [en] 2021-05-15T21:05:39Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_15&diff=148092 2006 AIME II Problems/Problem 15 2021-03-01T03:32:53Z <p>Math101010: /* Solution 1 (Geometric Interpretation) */</p> <hr /> <div>== Problem ==<br /> Given that &lt;math&gt; x, y, &lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; are [[real number]]s that satisfy:<br /> <br /> &lt;center&gt;&lt;math&gt; x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}} &lt;/math&gt; &lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt; y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}} &lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt; z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}&lt;/math&gt;&lt;/center&gt;<br /> <br /> and that &lt;math&gt; x+y+z = \frac{m}{\sqrt{n}}, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are positive integers and &lt;math&gt; n &lt;/math&gt; is not divisible by the square of any prime, find &lt;math&gt; m+n.&lt;/math&gt;<br /> <br /> == Solution 1 (Geometric Interpretation)==<br /> Let &lt;math&gt;\triangle XYZ&lt;/math&gt; be a triangle with sides of length &lt;math&gt;x, y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;, and suppose this triangle is acute (so all [[altitude]]s are in the interior of the triangle).<br /> <br /> <br /> Let the altitude to the side of length &lt;math&gt;x&lt;/math&gt; be of length &lt;math&gt;h_x&lt;/math&gt;, and similarly for &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. Then we have by two applications of the [[Pythagorean Theorem]] we that &lt;cmath&gt;x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}&lt;/cmath&gt; <br /> As a [[function]] of &lt;math&gt;h_x&lt;/math&gt;, the [[RHS]] of this equation is strictly decreasing, so it takes each value in its [[range]] exactly once. Thus we must have that &lt;math&gt;h_x^2 = \frac1{16}&lt;/math&gt; and so &lt;math&gt;h_x = \frac{1}4&lt;/math&gt; and similarly &lt;math&gt;h_y = \frac15&lt;/math&gt; and &lt;math&gt;h_z = \frac16&lt;/math&gt;.<br /> <br /> <br /> The area of the triangle must be the same no matter how we measure it; therefore &lt;math&gt;x\cdot h_x = y\cdot h_y = z \cdot h_z&lt;/math&gt; gives us &lt;math&gt;\frac x4 = \frac y5 = \frac z6 = 2A&lt;/math&gt; and &lt;math&gt;x = 8A, y = 10A&lt;/math&gt; and &lt;math&gt;z = 12A&lt;/math&gt;. <br /> <br /> <br /> The [[semiperimeter]] of the triangle is &lt;math&gt;s = \frac{8A + 10A + 12A}{2} = 15A&lt;/math&gt; so by [[Heron's formula]] we have &lt;cmath&gt;A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}&lt;/cmath&gt; <br /> <br /> Thus, &lt;math&gt;A = \frac{1}{15\sqrt{7}}&lt;/math&gt; and &lt;math&gt;x + y + z = 30A = \frac2{\sqrt{7}}&lt;/math&gt; and the answer is &lt;math&gt;2 + 7 = \boxed{009}&lt;/math&gt;.<br /> -------------------------------------<br /> The justification that there is an acute triangle with sides of length &lt;math&gt;x, y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;:<br /> <br /> Note that &lt;math&gt;x, y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; are each the sum of two positive [[square root]]s of real numbers, so &lt;math&gt;x, y, z \geq 0&lt;/math&gt;. (Recall that, by [[AIME]] [[mathematical convention| convention]], all numbers (including square roots) are taken to be real unless otherwise indicated.) <br /> <br /> Also, &lt;math&gt;\sqrt{y^2-\frac{1}{16}} &lt; \sqrt{y^2} = y&lt;/math&gt;, so we have &lt;math&gt;x &lt; y + z&lt;/math&gt;, &lt;math&gt;y &lt; z + x&lt;/math&gt; and &lt;math&gt;z &lt; x + y&lt;/math&gt;. But these conditions are exactly those of the [[triangle inequality]], so there does exist such a triangle.<br /> <br /> == Solution 2 (Algebraic) ==<br /> Note that none of &lt;math&gt;x,y,z&lt;/math&gt; can be zero.<br /> <br /> Each of the equations is in the form &lt;cmath&gt;a=\sqrt{b^2-d^2}+\sqrt{c^2-d^2}&lt;/cmath&gt;<br /> <br /> Isolate a radical and square the equation to get &lt;cmath&gt;b^2-d^2=a^2-2a\sqrt{c^2-d^2}+c^2-d^2&lt;/cmath&gt; <br /> <br /> Now cancel, and again isolate the radical, and square the equation to get &lt;cmath&gt;a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2&lt;/cmath&gt;<br /> <br /> Rearranging gives &lt;cmath&gt;a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2&lt;/cmath&gt;<br /> <br /> Now note that everything is cyclic but the last term (i.e. &lt;math&gt;-4a^2d^2&lt;/math&gt;), which implies &lt;cmath&gt;-4x^2\cdot\frac1{16}=-4y^2\cdot\frac1{25}=-4z^2\cdot\frac1{36}&lt;/cmath&gt;<br /> <br /> Or<br /> <br /> &lt;cmath&gt;x: y: z=4: 5: 6 \implies x=\frac{4y}5 \textrm{ and } z=\frac{6y}5&lt;/cmath&gt;<br /> <br /> Plug these values into the middle equation to get &lt;cmath&gt;\frac{256y^4+625y^4+1296y^4}{625}=\frac{800y^4}{625}+\frac{1800y^4}{625}+\frac{1152y^4}{625}-\frac{100y^2}{625}&lt;/cmath&gt;<br /> <br /> Simplifying gives &lt;cmath&gt;1575y^4=100y^2 \textrm{ but } y \neq 0 \implies y^2=\frac4{63} \textrm{ or } y=\frac2{3\sqrt7}&lt;/cmath&gt;<br /> <br /> Substituting the value of &lt;math&gt;y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; gives<br /> &lt;cmath&gt;x+y+z = \frac{4y+5y+6y}5 = 3y = 3 \cdot \frac{2}{3\sqrt7} = \frac{2}{\sqrt7}&lt;/cmath&gt; <br /> <br /> And thus the answer is &lt;math&gt;\boxed{009}&lt;/math&gt;<br /> <br /> ~phoenixfire<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=M6sC26dzb_I<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_14&diff=148091 2006 AIME II Problems/Problem 14 2021-03-01T03:32:09Z <p>Math101010: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; S_n &lt;/math&gt; be the sum of the [[reciprocal]]s of the non-zero digits of the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt; 10^n &lt;/math&gt; inclusive. Find the smallest positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt; S_n &lt;/math&gt; is an integer.<br /> <br /> == Solution ==<br /> Let &lt;math&gt;K = \sum_{i=1}^{9}{\frac{1}{i}}&lt;/math&gt;. Examining the terms in &lt;math&gt;S_1&lt;/math&gt;, we see that &lt;math&gt;S_1 = K + 1&lt;/math&gt; since each digit &lt;math&gt;n&lt;/math&gt; appears once and 1 appears an extra time. Now consider writing out &lt;math&gt;S_2&lt;/math&gt;. Each term of &lt;math&gt;K&lt;/math&gt; will appear 10 times in the units place and 10 times in the tens place (plus one extra 1 will appear), so &lt;math&gt;S_2 = 20K + 1&lt;/math&gt;.<br /> <br /> In general, we will have that<br /> &lt;center&gt;&lt;math&gt;S_n = (n10^{n-1})K + 1&lt;/math&gt;&lt;/center&gt;<br /> <br /> because each digit will appear &lt;math&gt;10^{n - 1}&lt;/math&gt; times in each place in the numbers &lt;math&gt;1, 2, \ldots, 10^{n} - 1&lt;/math&gt;, and there are &lt;math&gt;n&lt;/math&gt; total places. <br /> <br /> The denominator of &lt;math&gt;K&lt;/math&gt; is &lt;math&gt;D = 2^3\cdot 3^2\cdot 5\cdot 7&lt;/math&gt;. For &lt;math&gt;S_n&lt;/math&gt; to be an integer, &lt;math&gt;n10^{n-1}&lt;/math&gt; must be divisible by &lt;math&gt;D&lt;/math&gt;. Since &lt;math&gt;10^{n-1}&lt;/math&gt; only contains the factors &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; (but will contain enough of them when &lt;math&gt;n \geq 3&lt;/math&gt;), we must choose &lt;math&gt;n&lt;/math&gt; to be [[divisible]] by &lt;math&gt;3^2\cdot 7&lt;/math&gt;. Since we're looking for the smallest such &lt;math&gt;n&lt;/math&gt;, the answer is &lt;math&gt;\boxed{063}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=1960_IMO_Problems/Problem_1&diff=147220 1960 IMO Problems/Problem 1 2021-02-17T01:16:13Z <p>Math101010: /* Problem */ added &quot;digits of $N$&quot; at end of problem statement</p> <hr /> <div>==Problem==<br /> <br /> Determine all three-digit numbers &lt;math&gt;N&lt;/math&gt; having the property that &lt;math&gt;N&lt;/math&gt; is divisible by 11, and &lt;math&gt;\dfrac{N}{11}&lt;/math&gt; is equal to the sum of the squares of the digits of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> == Solutions ==<br /> ===Solution 1===<br /> Let &lt;math&gt;N = 100a + 10b+c&lt;/math&gt; for some digits &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. Then &lt;cmath&gt; 100a + 10b+c = 11m&lt;/cmath&gt; for some &lt;math&gt;m&lt;/math&gt;. We also have &lt;math&gt;m=a^2+b^2+c^2&lt;/math&gt;. Substituting this into the first equation and simplification, we get &lt;cmath&gt;100a+10b+c = 11a^2 +11b^2 +11c^2&lt;/cmath&gt; For an integer divisible by &lt;math&gt;11&lt;/math&gt;, the the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by &lt;math&gt;11&lt;/math&gt;. Thus we get: &lt;math&gt;b = a + c&lt;/math&gt; or &lt;math&gt;b = a + c - 11&lt;/math&gt;.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Let &lt;math&gt;b=a+c&lt;/math&gt;. We get &lt;cmath&gt;100a+c+10a+10c = 11a^2 +11c^2+11(a+c)^2&lt;/cmath&gt; &lt;cmath&gt;10a+c = 2a^2+2ac+2c^2&lt;/cmath&gt; Since the right side is even, the left side must also be even. Let &lt;math&gt;c=2q&lt;/math&gt; for some &lt;math&gt;q = 0,1,2,3,4&lt;/math&gt;. Then &lt;cmath&gt;10a+2q=2a^2+4aq+8q^2&lt;/cmath&gt;&lt;cmath&gt;5a+q=a^2+2aq+4q^2&lt;/cmath&gt; <br /> Substitute &lt;math&gt;q=0,1,2,3,4&lt;/math&gt; into the last equation and then solve for &lt;math&gt;a&lt;/math&gt;. <br /> <br /> When &lt;math&gt;q=0&lt;/math&gt;, we get &lt;math&gt;a=5&lt;/math&gt;. Thus &lt;math&gt;c=0&lt;/math&gt; and &lt;math&gt;b=5&lt;/math&gt;. We get that &lt;math&gt;N=550&lt;/math&gt; which works.<br /> <br /> When &lt;math&gt;q=1&lt;/math&gt;, we get that &lt;math&gt;a&lt;/math&gt; is not an integer. There is no &lt;math&gt;N&lt;/math&gt; for this case. <br /> <br /> When &lt;math&gt;q=2&lt;/math&gt;, we get that &lt;math&gt;a&lt;/math&gt; is not an integer. There is no &lt;math&gt;N&lt;/math&gt; for this case.<br /> <br /> When &lt;math&gt;q=3&lt;/math&gt;, we get that &lt;math&gt;a&lt;/math&gt; is not an integer. There is no &lt;math&gt;N&lt;/math&gt; for this case.<br /> <br /> When &lt;math&gt;q=4&lt;/math&gt;, we get that &lt;math&gt;a&lt;/math&gt; is not an integer. There is no &lt;math&gt;N&lt;/math&gt; for this case.<br /> <br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Let &lt;math&gt;b = a + c - 11&lt;/math&gt;. We get &lt;cmath&gt;100a+c+10a+10c -110= 11(a^2+(a+c)^2-22(a+c)+c^2+121)&lt;/cmath&gt; &lt;cmath&gt; 10a+c=2a^2+2c^2+2ac-22a-22c+131&lt;/cmath&gt; &lt;cmath&gt;2(a-8)^2+2(c-\frac{23}{4})^2+2ac-\frac{505}{8}=0&lt;/cmath&gt;<br /> Now we test all &lt;math&gt;c=0\rightarrow10&lt;/math&gt;. When &lt;math&gt;c=0,1,2,4,5,6,7,8,9&lt;/math&gt;, we get no integer solution to &lt;math&gt;a&lt;/math&gt;. Thus, for these values of &lt;math&gt;c&lt;/math&gt;, there is no valid &lt;math&gt;N&lt;/math&gt;. However, when &lt;math&gt;c=3&lt;/math&gt;, we get &lt;cmath&gt;2(a-8)^2+2(3-\frac{23}{4})^2+6a-\frac{505}{8}=0&lt;/cmath&gt; &lt;cmath&gt;2(a-8)^2+6a-48 = 0&lt;/cmath&gt; We get that &lt;math&gt;a=8&lt;/math&gt; is a valid solution. For this case, we get &lt;math&gt;a=8,b=0,c=3&lt;/math&gt;, so &lt;math&gt;N=803&lt;/math&gt;, and this is a valid value. Thus, the answers are &lt;math&gt;\boxed{N=550,803}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Define a '''ten''' to be all ten positive integers which begin with a fixed tens digit.<br /> <br /> We can make a systematic approach to this:<br /> <br /> <br /> By inspection, &lt;math&gt;\dfrac{N}{11}&lt;/math&gt; must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well.<br /> <br /> For a given ten, the sum of the squares of the digits of &lt;math&gt;N&lt;/math&gt; increases faster than &lt;math&gt;\dfrac{N}{11}&lt;/math&gt;, so we can have at most one number in every ten that works.<br /> <br /> We check the first ten:<br /> <br /> &lt;math&gt;11*11=121&lt;/math&gt;<br /> <br /> &lt;math&gt;1^2+2^2+1^2=4&lt;/math&gt;<br /> <br /> &lt;math&gt;12*11=132&lt;/math&gt;<br /> <br /> &lt;math&gt;1^2+3^2+2^2=14&lt;/math&gt;<br /> <br /> 11 is too small and 12 is too large, so all numbers below 11 will be too small and all numbers above 12 will be too large, so no numbers in the first ten work.<br /> <br /> We try the second ten:<br /> <br /> &lt;math&gt;21*11=231&lt;/math&gt;<br /> <br /> &lt;math&gt;2^2+3^2+1^2=14&lt;/math&gt;<br /> <br /> &lt;math&gt;22*11=242&lt;/math&gt;<br /> <br /> &lt;math&gt;2^2+4^2+2^2=24&lt;/math&gt;<br /> <br /> Therefore, no numbers in the second ten work.<br /> <br /> We continue, to find out that 50 and 73 are the only ones that works.<br /> <br /> &lt;math&gt;N=50*11=550&lt;/math&gt;, &lt;math&gt;N=73*11=803&lt;/math&gt; so there are two &lt;math&gt;N&lt;/math&gt; that works.<br /> <br /> ===Solution 3===<br /> <br /> Note that there are only 900 three-digit numbers. Of these, only 81 are divisible by 11. This brings to mind the great strategy of listing and bashing. Being organized while listing will get you the answer. There is plenty of time on the IMO, and listing is highly unlikely to miss cases. We end up getting &lt;math&gt;\boxed{N=550,803}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{IMO7 box|year=1960|before=First Question|num-a=2}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_13&diff=145851 2021 AMC 12B Problems/Problem 13 2021-02-12T06:52:29Z <p>Math101010: /* Problem */</p> <hr /> <div>==Problem==<br /> How many values of &lt;math&gt;\theta&lt;/math&gt; in the interval &lt;math&gt;0&lt;\theta\le 2\pi&lt;/math&gt; satisfy&lt;cmath&gt;1-3\sin\theta+5\cos3\theta = 0?&lt;/cmath&gt;&lt;math&gt;\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> First, move terms to get &lt;math&gt;1+5cos3x=3sinx&lt;/math&gt;. After graphing, we find that there are &lt;math&gt;\boxed{6}&lt;/math&gt; solutions (two in each period of &lt;math&gt;5cos3x&lt;/math&gt;). -dstanz5<br /> <br /> <br /> == Video Solution by OmegaLearn (Using Sine and Cosine Graph) ==<br /> https://youtu.be/toBOpc6vS6s<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems&diff=141833 2020 AMC 10A Problems 2021-01-10T07:18:24Z <p>Math101010: Undo revision 141826 by Wadehua (talk)</p> <hr /> <div>{{AMC10 Problems|year=2020|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> What value of &lt;math&gt;x&lt;/math&gt; satisfies<br /> <br /> &lt;cmath&gt;x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ -\frac{2}{3}\qquad\textbf{(B)}\ \frac{7}{36}\qquad\textbf{(C)}\ \frac{7}{12}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{5}{6}&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> The numbers &lt;math&gt;3, 5, 7, a,&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; have an average (arithmetic mean) of &lt;math&gt;15&lt;/math&gt;. What is the average of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> Assuming &lt;math&gt;a\neq3&lt;/math&gt;, &lt;math&gt;b\neq4&lt;/math&gt;, and &lt;math&gt;c\neq5&lt;/math&gt;, what is the value in simplest form of the following expression?<br /> &lt;cmath&gt;\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } -1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) } \frac{1}{60} - \frac{1}{abc}&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A driver travels for &lt;math&gt;2&lt;/math&gt; hours at &lt;math&gt;60&lt;/math&gt; miles per hour, during which her car gets &lt;math&gt;30&lt;/math&gt; miles per gallon of gasoline. She is paid &lt;math&gt;\$0.50&lt;/math&gt; per mile, and her only expense is gasoline at &lt;math&gt;\$2.00&lt;/math&gt; per gallon. What is her net rate of pay, in dollars per hour, after this expense?<br /> <br /> &lt;math&gt;\textbf{(A) }20 \qquad\textbf{(B) }22 \qquad\textbf{(C) }24 \qquad\textbf{(D) } 25\qquad\textbf{(E) } 26&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> What is the sum of all real numbers &lt;math&gt;x&lt;/math&gt; for which &lt;math&gt;|x^2-12x+34|=2?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> How many &lt;math&gt;4&lt;/math&gt;-digit positive integers (that is, integers between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt;, inclusive) having only even digits are divisible by &lt;math&gt;5?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> The &lt;math&gt;25&lt;/math&gt; integers from &lt;math&gt;-10&lt;/math&gt; to &lt;math&gt;14,&lt;/math&gt; inclusive, can be arranged to form a &lt;math&gt;5&lt;/math&gt;-by-&lt;math&gt;5&lt;/math&gt; square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?<br /> <br /> &lt;math&gt;\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of<br /> <br /> &lt;cmath&gt;1+2+3-4+5+6+7-8+\cdots+197+198+199-200?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> A single bench section at a school event can hold either &lt;math&gt;7&lt;/math&gt; adults or &lt;math&gt;11&lt;/math&gt; children. When &lt;math&gt;N&lt;/math&gt; bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Seven cubes, whose volumes are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;27&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, &lt;math&gt;125&lt;/math&gt;, &lt;math&gt;216&lt;/math&gt;, and &lt;math&gt;343&lt;/math&gt; cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> What is the median of the following list of &lt;math&gt;4040&lt;/math&gt; numbers&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;cmath&gt;1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Triangle &lt;math&gt;AMC&lt;/math&gt; is isosceles with &lt;math&gt;AM = AC&lt;/math&gt;. Medians &lt;math&gt;\overline{MV}&lt;/math&gt; and &lt;math&gt;\overline{CU}&lt;/math&gt; are perpendicular to each other, and &lt;math&gt;MV=CU=12&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AMC?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((-4,0)--(4,0)--(0,12)--cycle);<br /> draw((-2,6)--(4,0));<br /> draw((2,6)--(-4,0));<br /> label(&quot;M&quot;, (-4,0), W);<br /> label(&quot;C&quot;, (4,0), E);<br /> label(&quot;A&quot;, (0, 12), N);<br /> label(&quot;V&quot;, (2, 6), NE);<br /> label(&quot;U&quot;, (-2, 6), NW);<br /> label(&quot;P&quot;, (0, 3.6), S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of&lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;&lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A positive integer divisor of &lt;math&gt;12!&lt;/math&gt; is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A point is chosen at random within the square in the coordinate plane whose vertices are &lt;math&gt;(0, 0), (2020, 0), (2020, 2020),&lt;/math&gt; and &lt;math&gt;(0, 2020)&lt;/math&gt;. The probability that the point is within &lt;math&gt;d&lt;/math&gt; units of a lattice point is &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;. (A point &lt;math&gt;(x, y)&lt;/math&gt; is a lattice point if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both integers.) What is &lt;math&gt;d&lt;/math&gt; to the nearest tenth?<br /> <br /> &lt;math&gt;\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Define&lt;cmath&gt;P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).&lt;/cmath&gt;How many integers &lt;math&gt;n&lt;/math&gt; are there such that &lt;math&gt;P(n)\leq 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Let &lt;math&gt;(a,b,c,d)&lt;/math&gt; be an ordered quadruple of not necessarily distinct integers, each one of them in the set &lt;math&gt;{0,1,2,3}.&lt;/math&gt; For how many such quadruples is it true that &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; is odd? (For example, &lt;math&gt;(0,3,1,1)&lt;/math&gt; is one such quadruple, because &lt;math&gt;0\cdot 1-3\cdot 1 = -3&lt;/math&gt; is odd.)<br /> <br /> &lt;math&gt;\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> As shown in the figure below, a regular dodecahedron (the polyhedron consisting of &lt;math&gt;12&lt;/math&gt; congruent regular pentagonal faces) floats in empty space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> unitsize(5cm);<br /> pair A = (0.082, 0.378);<br /> pair B = (0.091, 0.649);<br /> pair C = (0.249, 0.899);<br /> pair D = (0.479, 0.939);<br /> pair E = (0.758, 0.893);<br /> pair F = (0.862, 0.658);<br /> pair G = (0.924, 0.403);<br /> pair H = (0.747, 0.194);<br /> pair I = (0.526, 0.075);<br /> pair J = (0.251, 0.170);<br /> pair K = (0.568, 0.234);<br /> pair L = (0.262, 0.449);<br /> pair M = (0.373, 0.813);<br /> pair N = (0.731, 0.813);<br /> pair O = (0.851, 0.461);<br /> path[] f;<br /> f = A--B--C--M--L--cycle;<br /> f = C--D--E--N--M--cycle;<br /> f = E--F--G--O--N--cycle;<br /> f = G--H--I--K--O--cycle;<br /> f = I--J--A--L--K--cycle;<br /> f = K--L--M--N--O--cycle;<br /> draw(f);<br /> axialshade(f, white, M, gray(0.5), (C+2*D)/3);<br /> draw(f);<br /> filldraw(f, gray);<br /> filldraw(f, gray);<br /> axialshade(f, white, L, gray(0.7), J);<br /> draw(f);<br /> draw(f);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> There exists a unique strictly increasing sequence of nonnegative integers &lt;math&gt;a_1 &lt; a_2 &lt; … &lt; a_k&lt;/math&gt; such that&lt;cmath&gt;\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.&lt;/cmath&gt;What is &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> For how many positive integers &lt;math&gt;n \le 1000&lt;/math&gt; is&lt;cmath&gt;\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor&lt;/cmath&gt;not divisible by &lt;math&gt;3&lt;/math&gt;? (Recall that &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Let &lt;math&gt;T&lt;/math&gt; be the triangle in the coordinate plane with vertices &lt;math&gt;(0,0), (4,0),&lt;/math&gt; and &lt;math&gt;(0,3).&lt;/math&gt; Consider the following five isometries (rigid transformations) of the plane: rotations of &lt;math&gt;90^{\circ}, 180^{\circ},&lt;/math&gt; and &lt;math&gt;270^{\circ}&lt;/math&gt; counterclockwise around the origin, reflection across the &lt;math&gt;x&lt;/math&gt;-axis, and reflection across the &lt;math&gt;y&lt;/math&gt;-axis. How many of the &lt;math&gt;125&lt;/math&gt; sequences of three of these transformations (not necessarily distinct) will return &lt;math&gt;T&lt;/math&gt; to its original position? (For example, a &lt;math&gt;180^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by a reflection across the &lt;math&gt;y&lt;/math&gt;-axis will return &lt;math&gt;T&lt;/math&gt; to its original position, but a &lt;math&gt;90^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by another reflection across the &lt;math&gt;x&lt;/math&gt;-axis will not return &lt;math&gt;T&lt;/math&gt; to its original position.)<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the least positive integer greater than &lt;math&gt;1000&lt;/math&gt; for which&lt;cmath&gt;\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.&lt;/cmath&gt;What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly &lt;math&gt;7.&lt;/math&gt; Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{7}{36} \qquad\textbf{(B) } \frac{5}{24} \qquad\textbf{(C) } \frac{2}{9} \qquad\textbf{(D) } \frac{17}{72} \qquad\textbf{(E) } \frac{1}{4}&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2020|ab=A|before=[[2019 AMC 10B Problems]]|after=[[2020 AMC 10B Problems]]}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems&diff=141832 2020 AMC 10A Problems 2021-01-10T07:18:09Z <p>Math101010: Undo revision 141831 by Wadehua (talk)</p> <hr /> <div>{{AMC10 Problems|year=2020|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> What value of &lt;math&gt;x&lt;/math&gt; satisfies<br /> <br /> &lt;cmath&gt;x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ -\frac{2}{3}\qquad\textbf{(B)}\ \frac{7}{36}\qquad\textbf{(C)}\ \frac{7}{12}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{5}{6}&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> The numbers &lt;math&gt;3, 5, 7, a,&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; have an average (arithmetic mean) of &lt;math&gt;15&lt;/math&gt;. What is the average of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> Assuming &lt;math&gt;a\neq3&lt;/math&gt;, &lt;math&gt;b\neq4&lt;/math&gt;, and &lt;math&gt;c\neq5&lt;/math&gt;, what is the value in simplest form of the following expression?<br /> &lt;cmath&gt;\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } -1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) } \frac{1}{60} - \frac{1}{abc}&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A driver travels for &lt;math&gt;2&lt;/math&gt; hours at &lt;math&gt;60&lt;/math&gt; miles per hour, during which her car gets &lt;math&gt;30&lt;/math&gt; miles per gallon of gasoline. She is paid &lt;math&gt;\$0.50&lt;/math&gt; per mile, and her only expense is gasoline at &lt;math&gt;\$2.00&lt;/math&gt; per gallon. What is her net rate of pay, in dollars per hour, after this expense?<br /> <br /> &lt;math&gt;\textbf{(A) }20 \qquad\textbf{(B) }22 \qquad\textbf{(C) }24 \qquad\textbf{(D) } 25\qquad\textbf{(E) } 26&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> What is the sum of all real numbers &lt;math&gt;x&lt;/math&gt; for which &lt;math&gt;|x^2-12x+34|=2?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> How many &lt;math&gt;4&lt;/math&gt;-digit positive integers (that is, integers between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt;, inclusive) having only even digits are divisible by &lt;math&gt;5?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> The &lt;math&gt;25&lt;/math&gt; integers from &lt;math&gt;-10&lt;/math&gt; to &lt;math&gt;14,&lt;/math&gt; inclusive, can be arranged to form a &lt;math&gt;5&lt;/math&gt;-by-&lt;math&gt;5&lt;/math&gt; square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?<br /> <br /> &lt;math&gt;\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of<br /> <br /> &lt;cmath&gt;1+2+3-4+5+6+7-8+\cdots+197+198+199-200?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> A single bench section at a school event can hold either &lt;math&gt;7&lt;/math&gt; adults or &lt;math&gt;11&lt;/math&gt; children. When &lt;math&gt;N&lt;/math&gt; bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Seven cubes, whose volumes are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;27&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, &lt;math&gt;125&lt;/math&gt;, &lt;math&gt;216&lt;/math&gt;, and &lt;math&gt;343&lt;/math&gt; cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> What is the median of the following list of &lt;math&gt;4040&lt;/math&gt; numbers&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;cmath&gt;1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> math is stupid<br /> <br /> ==Problem 13==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of&lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;&lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A positive integer divisor of &lt;math&gt;12!&lt;/math&gt; is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A point is chosen at random within the square in the coordinate plane whose vertices are &lt;math&gt;(0, 0), (2020, 0), (2020, 2020),&lt;/math&gt; and &lt;math&gt;(0, 2020)&lt;/math&gt;. The probability that the point is within &lt;math&gt;d&lt;/math&gt; units of a lattice point is &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;. (A point &lt;math&gt;(x, y)&lt;/math&gt; is a lattice point if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both integers.) What is &lt;math&gt;d&lt;/math&gt; to the nearest tenth?<br /> <br /> &lt;math&gt;\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Define&lt;cmath&gt;P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).&lt;/cmath&gt;How many integers &lt;math&gt;n&lt;/math&gt; are there such that &lt;math&gt;P(n)\leq 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Let &lt;math&gt;(a,b,c,d)&lt;/math&gt; be an ordered quadruple of not necessarily distinct integers, each one of them in the set &lt;math&gt;{0,1,2,3}.&lt;/math&gt; For how many such quadruples is it true that &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; is odd? (For example, &lt;math&gt;(0,3,1,1)&lt;/math&gt; is one such quadruple, because &lt;math&gt;0\cdot 1-3\cdot 1 = -3&lt;/math&gt; is odd.)<br /> <br /> &lt;math&gt;\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> As shown in the figure below, a regular dodecahedron (the polyhedron consisting of &lt;math&gt;12&lt;/math&gt; congruent regular pentagonal faces) floats in empty space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> unitsize(5cm);<br /> pair A = (0.082, 0.378);<br /> pair B = (0.091, 0.649);<br /> pair C = (0.249, 0.899);<br /> pair D = (0.479, 0.939);<br /> pair E = (0.758, 0.893);<br /> pair F = (0.862, 0.658);<br /> pair G = (0.924, 0.403);<br /> pair H = (0.747, 0.194);<br /> pair I = (0.526, 0.075);<br /> pair J = (0.251, 0.170);<br /> pair K = (0.568, 0.234);<br /> pair L = (0.262, 0.449);<br /> pair M = (0.373, 0.813);<br /> pair N = (0.731, 0.813);<br /> pair O = (0.851, 0.461);<br /> path[] f;<br /> f = A--B--C--M--L--cycle;<br /> f = C--D--E--N--M--cycle;<br /> f = E--F--G--O--N--cycle;<br /> f = G--H--I--K--O--cycle;<br /> f = I--J--A--L--K--cycle;<br /> f = K--L--M--N--O--cycle;<br /> draw(f);<br /> axialshade(f, white, M, gray(0.5), (C+2*D)/3);<br /> draw(f);<br /> filldraw(f, gray);<br /> filldraw(f, gray);<br /> axialshade(f, white, L, gray(0.7), J);<br /> draw(f);<br /> draw(f);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> There exists a unique strictly increasing sequence of nonnegative integers &lt;math&gt;a_1 &lt; a_2 &lt; … &lt; a_k&lt;/math&gt; such that&lt;cmath&gt;\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.&lt;/cmath&gt;What is &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> For how many positive integers &lt;math&gt;n \le 1000&lt;/math&gt; is&lt;cmath&gt;\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor&lt;/cmath&gt;not divisible by &lt;math&gt;3&lt;/math&gt;? (Recall that &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Let &lt;math&gt;T&lt;/math&gt; be the triangle in the coordinate plane with vertices &lt;math&gt;(0,0), (4,0),&lt;/math&gt; and &lt;math&gt;(0,3).&lt;/math&gt; Consider the following five isometries (rigid transformations) of the plane: rotations of &lt;math&gt;90^{\circ}, 180^{\circ},&lt;/math&gt; and &lt;math&gt;270^{\circ}&lt;/math&gt; counterclockwise around the origin, reflection across the &lt;math&gt;x&lt;/math&gt;-axis, and reflection across the &lt;math&gt;y&lt;/math&gt;-axis. How many of the &lt;math&gt;125&lt;/math&gt; sequences of three of these transformations (not necessarily distinct) will return &lt;math&gt;T&lt;/math&gt; to its original position? (For example, a &lt;math&gt;180^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by a reflection across the &lt;math&gt;y&lt;/math&gt;-axis will return &lt;math&gt;T&lt;/math&gt; to its original position, but a &lt;math&gt;90^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by another reflection across the &lt;math&gt;x&lt;/math&gt;-axis will not return &lt;math&gt;T&lt;/math&gt; to its original position.)<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the least positive integer greater than &lt;math&gt;1000&lt;/math&gt; for which&lt;cmath&gt;\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.&lt;/cmath&gt;What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly &lt;math&gt;7.&lt;/math&gt; Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{7}{36} \qquad\textbf{(B) } \frac{5}{24} \qquad\textbf{(C) } \frac{2}{9} \qquad\textbf{(D) } \frac{17}{72} \qquad\textbf{(E) } \frac{1}{4}&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2020|ab=A|before=[[2019 AMC 10B Problems]]|after=[[2020 AMC 10B Problems]]}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_16&diff=104009 2012 AMC 10A Problems/Problem 16 2019-03-03T21:41:39Z <p>Math101010: /* Problem */</p> <hr /> <div>==Problem ==<br /> <br /> Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000 &lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let &lt;math&gt;x&lt;/math&gt; be the time these runners run in seconds.<br /> <br /> &lt;cmath&gt;4.8x-4.4x=500 \Rightarrow x=1250&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;4.4(1250)=5500&lt;/math&gt; is a multiple of 500, it turns out they just meet back at the start line.<br /> <br /> Now we must find a time that is a multiple of &lt;math&gt;1250&lt;/math&gt; and results in the 5.0 m/s runner to end up on the start line. Every &lt;math&gt;1250&lt;/math&gt; seconds, that fastest runner goes &lt;math&gt;5.0(1250)=6250&lt;/math&gt; meters. In &lt;math&gt;2(1250)=2500&lt;/math&gt; seconds, he goes &lt;math&gt;5.0(2500)=12500&lt;/math&gt; meters. Therefore the runners run &lt;math&gt;\boxed{\textbf{(C)}\ 2,500}&lt;/math&gt; seconds.<br /> <br /> == Solution 2==<br /> <br /> Working backwards from the answers starting with the smallest answer, if they had run &lt;math&gt;1000&lt;/math&gt; seconds, they would have run &lt;math&gt;4400, 4800, 5000&lt;/math&gt; meters, respectively. The first two runners have a difference of &lt;math&gt;400&lt;/math&gt; meters, which is not a multiple of &lt;math&gt;500&lt;/math&gt; (one lap), so they are not in the same place.<br /> <br /> If they had run &lt;math&gt;1250&lt;/math&gt; seconds, the runners would have run &lt;math&gt;5500, 6000, 6250&lt;/math&gt; meters, respectively. The last two runners have a difference of &lt;math&gt;250&lt;/math&gt; meters, which is not a multiple of &lt;math&gt;500&lt;/math&gt;.<br /> <br /> If they had run &lt;math&gt;2500&lt;/math&gt; seconds, the runners would have run &lt;math&gt;11000, 12000, 12500&lt;/math&gt; meters, respectively. The distance separating each pair of runners is a multiple of &lt;math&gt;500&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C)}\ 2,500}&lt;/math&gt; seconds.<br /> <br /> == Solution 3==<br /> <br /> Let &lt;math&gt;t&lt;/math&gt; be the time run in seconds, then the difference in meters run between the three runners is &lt;math&gt;0.2t, 0.4t, 0.6t&lt;/math&gt;. For them to be at the same location all of them need to be multiples of 500. It is now easy to see that &lt;math&gt;0.2t=500, 0.4t=1000, 0.6t=1500&lt;/math&gt;, so &lt;math&gt;t=\boxed{\textbf{(C)}\ 2,500}&lt;/math&gt;.<br /> <br /> == Solution 4==<br /> After &lt;math&gt;t&lt;/math&gt; seconds, respectively the runners would've ran &lt;math&gt;4.4t, 4.8t,&lt;/math&gt; and &lt;math&gt;5t&lt;/math&gt; meters. Their current positions on the track are these values &lt;math&gt;\pmod{500}&lt;/math&gt;. We're trying to find the value of &lt;math&gt;t&lt;/math&gt; such that &lt;cmath&gt;4.4t \equiv 4.8t \equiv 5t \pmod{500}&lt;/cmath&gt; Subtracting &lt;math&gt;4.4t&lt;/math&gt; on all sides, we get &lt;cmath&gt;0 \equiv 0.4t \equiv 0.6t \pmod{500}&lt;/cmath&gt; Now, we must find a value for &lt;math&gt;t&lt;/math&gt; such that both &lt;math&gt;0.6t&lt;/math&gt; and &lt;math&gt;0.4t&lt;/math&gt; are simultaneously multiples of &lt;math&gt;500&lt;/math&gt;.<br /> <br /> Plugging in &lt;math&gt;500&lt;/math&gt; for &lt;math&gt;0.4t&lt;/math&gt; we get &lt;math&gt;t=1250&lt;/math&gt;, but this does not work for &lt;math&gt;0.6t&lt;/math&gt; (&lt;math&gt;750&lt;/math&gt; isn't a multiple of &lt;math&gt;500&lt;/math&gt;). Plugging in &lt;math&gt;0.4t=1000&lt;/math&gt;, we get &lt;math&gt;t=2500&lt;/math&gt;, and this does work for &lt;math&gt;0.6t&lt;/math&gt;. <br /> <br /> Therefore, &lt;math&gt;t=2500&lt;/math&gt; and the answer is &lt;math&gt;\textbf{(C) } 2500&lt;/math&gt;.<br /> <br /> [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 15:21, 19 February 2016 (EST)<br /> <br /> *Note: Modular Arithmetic works only for integral values, so my usage of decimals is technically incorrect but the intuition leads to the right answer<br /> <br /> ==Solution 5==<br /> Similar to the solution above, but is much quicker and does not involve trial and error. This uses decimal mod arithmetic, which can be justified by intuition...<br /> After &lt;math&gt;t&lt;/math&gt; seconds, respectively the runners would've ran &lt;math&gt;4.4t, 4.8t,&lt;/math&gt; and &lt;math&gt;5t&lt;/math&gt; meters. These three values are congruent &lt;math&gt;\pmod{500}&lt;/math&gt;, so <br /> &lt;cmath&gt;4.4t \equiv 4.8t \equiv 5t \pmod{500}&lt;/cmath&gt;. Subtract 4.4t from all three sides to get 0, 0.4t, and 0.6t are congruent. Now all we need to find is a value of t for which 0.4t and 0.6t are congruent mod 500. Subtract 0.4t from both sides to get 0.2t and 0 are congruent mod 500, or that 0.2t=t/5 is a multiple of 500. Let t=500k, so we want 100k to be a multiple of 500, or k to be a multiple of 5. Therefore, the smallest value of t is when k=5, and when t=500k=500(5)=&lt;math&gt;2500 \space (\text{C})&lt;/math&gt;<br /> <br /> - Solution by mathchampion1<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems&diff=98530 1983 AIME Problems 2018-11-07T03:52:49Z <p>Math101010: </p> <hr /> <div>{{AIME Problems|year=1983}}<br /> <br /> == Problem 1 ==<br /> Let &lt;math&gt;x&lt;/math&gt;,&lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt;, and let &lt;math&gt;w&lt;/math&gt; be a positive number such that &lt;math&gt;\log_xw=24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt;, and &lt;math&gt;\log_{xyz}w=12&lt;/math&gt;. Find &lt;math&gt;\log_zw&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Let &lt;math&gt;f(x)=|x-p|+|x-15|+|x-p-15|&lt;/math&gt;, where &lt;math&gt;0 &lt; p &lt; 15&lt;/math&gt;. Determine the [[minimum]] value taken by &lt;math&gt;f(x)&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the [[interval]] &lt;math&gt;p \leq x\leq15&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> What is the product of the real roots of the equation &lt;math&gt;x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}&lt;/math&gt;?<br /> <br /> [[1983 AIME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is &lt;math&gt;\sqrt{50}&lt;/math&gt; cm, the length of &lt;math&gt;AB&lt;/math&gt; is 6 cm, and that of &lt;math&gt;BC&lt;/math&gt; is 2 cm. The angle &lt;math&gt;ABC&lt;/math&gt; is a right angle. Find the square of the distance (in centimeters) from &lt;math&gt;B&lt;/math&gt; to the center of the circle.<br /> <br /> &lt;asy&gt;<br /> size(150); defaultpen(linewidth(0.6)+fontsize(11));<br /> real r=10;<br /> pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;<br /> path P=circle(O,r);<br /> C=intersectionpoint(B--(B.x+r,B.y),P);<br /> draw(P);<br /> draw(C--B--A);<br /> dot(O); dot(A); dot(B); dot(C);<br /> label(&quot;$O$&quot;,O,SW);<br /> label(&quot;$A$&quot;,A,NE);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,SE);&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> Suppose that the sum of the squares of two complex numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;7&lt;/math&gt; and the sum of the cubes is &lt;math&gt;10&lt;/math&gt;. What is the largest real value that &lt;math&gt;x + y&lt;/math&gt; can have?<br /> <br /> [[1983 AIME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Let &lt;math&gt;a_n&lt;/math&gt; equal &lt;math&gt;6^{n}+8^{n}&lt;/math&gt;. Determine the remainder upon dividing &lt;math&gt;a_ {83}&lt;/math&gt; by &lt;math&gt;49&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let &lt;math&gt;P&lt;/math&gt; be the probability that at least two of the three had been sitting next to each other. If &lt;math&gt;P&lt;/math&gt; is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?<br /> <br /> [[1983 AIME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> What is the largest 2-digit prime factor of the integer &lt;math&gt;{200\choose 100}&lt;/math&gt;?<br /> <br /> [[1983 AIME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Find the minimum value of &lt;math&gt;\frac{9x^2\sin^2 x + 4}{x\sin x}&lt;/math&gt; for &lt;math&gt;0 &lt; x &lt; \pi&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The numbers &lt;math&gt;1447&lt;/math&gt;, &lt;math&gt;1005&lt;/math&gt;, and &lt;math&gt;1231&lt;/math&gt; have something in common. Each is a four-digit number beginning with &lt;math&gt;1&lt;/math&gt; that has exactly two identical digits. How many such numbers are there?<br /> <br /> [[1983 AIME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The solid shown has a square base of side length &lt;math&gt;s&lt;/math&gt;. The upper edge is parallel to the base and has length &lt;math&gt;2s&lt;/math&gt;. All other edges have length &lt;math&gt;s&lt;/math&gt;. Given that &lt;math&gt;s=6\sqrt{2}&lt;/math&gt;, what is the volume of the solid?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> size(170);<br /> pathpen = black+linewidth(0.65);<br /> pointpen = black;<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C); <br /> draw(E--F);<br /> label(&quot;A&quot;,A, S);<br /> label(&quot;B&quot;,B, S);<br /> label(&quot;C&quot;,C, S);<br /> label(&quot;D&quot;,D, S);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The length of diameter &lt;math&gt;AB&lt;/math&gt; is a two digit integer. Reversing the digits gives the length of a perpendicular chord &lt;math&gt;CD&lt;/math&gt;. The distance from their intersection point &lt;math&gt;H&lt;/math&gt; to the center &lt;math&gt;O&lt;/math&gt; is a positive rational number. Determine the length of &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> &lt;asy&gt;pointpen=black; pathpen=black+linewidth(0.65);<br /> pair O=(0,0),A=(-65/2,0),B=(65/2,0);<br /> pair H=(-((65/2)^2-28^2)^.5,0),C=(H.x,28),D=(H.x,-28);<br /> D(CP(O,A));D(MP(&quot;A&quot;,A,W)--MP(&quot;B&quot;,B,E));D(MP(&quot;C&quot;,C,N)--MP(&quot;D&quot;,D));<br /> dot(MP(&quot;H&quot;,H,SE));dot(MP(&quot;O&quot;,O,SE));&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> For &lt;math&gt;\{1, 2, 3, \ldots, n\}&lt;/math&gt; and each of its non-empty subsets, an alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example, the alternating sum for &lt;math&gt;\{1, 2, 3, 6,9\}&lt;/math&gt; is &lt;math&gt;9-6+3-2+1=5&lt;/math&gt; and for &lt;math&gt;\{5\}&lt;/math&gt; it is simply &lt;math&gt;5&lt;/math&gt;. Find the sum of all such alternating sums for &lt;math&gt;n=7&lt;/math&gt;.<br /> <br /> [[1983 AIME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> In the adjoining figure, two circles with radii &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are drawn with their centers &lt;math&gt;12&lt;/math&gt; units apart. At &lt;math&gt;P&lt;/math&gt;, one of the points of intersection, a line is drawn in such a way that the chords &lt;math&gt;QP&lt;/math&gt; and &lt;math&gt;PR&lt;/math&gt; have equal length. (&lt;math&gt;P&lt;/math&gt; is the midpoint of &lt;math&gt;QR&lt;/math&gt;) Find the square of the length of &lt;math&gt;QP&lt;/math&gt;. <br /> <br /> &lt;!-- [[Image:1983_AIME-14.png]] --&gt;<br /> &lt;asy&gt;size(160);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=3;<br /> pair O1=(0,0), O2=(12,0);<br /> path C1=Circle(O1,8), C2=Circle(O2,6);<br /> pair P=intersectionpoints(C1,C2);<br /> path C3=Circle(P,sqrt(130));<br /> pair Q=intersectionpoints(C3,C1);<br /> pair R=intersectionpoints(C3,C2);<br /> draw(C1);<br /> draw(C2);<br /> draw(O2--O1);<br /> dot(O1);<br /> dot(O2);<br /> draw(Q--R);<br /> label(&quot;$Q$&quot;,Q,NW);<br /> label(&quot;$P$&quot;,P,1.5*dir(80));<br /> label(&quot;$R$&quot;,R,NE);<br /> label(&quot;12&quot;,waypoint(O1--O2,0.4),S);&lt;/asy&gt;<br /> <br /> [[1983 AIME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> The adjoining figure shows two intersecting chords in a circle, with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the sine of the minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this fraction is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> &lt;asy&gt;size(140);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=1;<br /> pair O1=(0,0);<br /> pair A=(-0.91,-0.41);<br /> pair B=(-0.99,0.13);<br /> pair C=(0.688,0.728);<br /> pair D=(-0.25,0.97);<br /> path C1=Circle(O1,1);<br /> draw(C1);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> draw(A--D);<br /> draw(B--C);<br /> pair F=intersectionpoint(A--D,B--C);<br /> add(pathticks(A--F,1,0.5,0,3.5));<br /> add(pathticks(F--D,1,0.5,0,3.5));<br /> &lt;/asy&gt;<br /> &lt;!-- [[Image:1983_AIME-15.png]] --&gt;<br /> <br /> [[1983 AIME Problems/Problem 15|Solution]]<br /> <br /> == See Also ==<br /> <br /> {{AIME box|year=1983|before=First AIME|after=[[1984 AIME Problems]]}}<br /> <br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{MAA Notice}}<br /> [[Category:AIME Problems]]</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_25&diff=93829 2008 AMC 12B Problems/Problem 25 2018-04-06T20:16:57Z <p>Math101010: </p> <hr /> <div>==Problem 25==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a trapezoid with &lt;math&gt;AB||CD, AB=11, BC=5, CD=19,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Bisectors of &lt;math&gt;\angle A&lt;/math&gt; and &lt;math&gt;\angle D&lt;/math&gt; meet at &lt;math&gt;P&lt;/math&gt;, and bisectors of &lt;math&gt;\angle B&lt;/math&gt; and &lt;math&gt;\angle C&lt;/math&gt; meet at &lt;math&gt;Q&lt;/math&gt;. What is the area of hexagon &lt;math&gt;ABQCDP&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \textbf{(E)}\ 36\sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;center&gt;[[File:2008_AMC_12B_25.jpg‎]]&lt;/center&gt;<br /> <br /> Note: In the image AB and CD have been swapped from their given lengths in the problem. However, this doesn't affect any of the solving.<br /> <br /> Drop perpendiculars to &lt;math&gt;CD&lt;/math&gt; from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, and call the intersections &lt;math&gt;X,Y&lt;/math&gt; respectively. Now, &lt;math&gt;DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2&lt;/math&gt; and &lt;math&gt;DX+CY=19-11=8&lt;/math&gt;. Thus, &lt;math&gt;DX-CY=3&lt;/math&gt;.<br /> We conclude &lt;math&gt;DX=\frac{11}{2}&lt;/math&gt; and &lt;math&gt;CY=\frac{5}{2}&lt;/math&gt;.<br /> To simplify things even more, notice that &lt;math&gt;90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD&lt;/math&gt;, so &lt;math&gt;\angle P=\angle Q=90^{\circ}&lt;/math&gt;.<br /> <br /> Also, &lt;cmath&gt;\sin(\angle PAD)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}&lt;/cmath&gt;<br /> So the area of &lt;math&gt;\triangle APD&lt;/math&gt; is: &lt;cmath&gt;R\cdot c\sin a\sin b =\frac{7\cdot7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}&lt;/cmath&gt;<br /> <br /> Over to the other side: &lt;math&gt;\triangle BCY&lt;/math&gt; is &lt;math&gt;30-60-90&lt;/math&gt;, and is therefore congruent to &lt;math&gt;\triangle BCQ&lt;/math&gt;. So &lt;math&gt;[BCQ]=\frac{5\cdot5\sqrt{3}}{8}&lt;/math&gt;.<br /> <br /> The area of the hexagon is clearly &lt;math&gt;[ABCD]-([BCQ]+[APD])&lt;/math&gt;&lt;cmath&gt;=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}=30\sqrt{3}\implies\boxed{B}&lt;/cmath&gt;<br /> <br /> ==Alternate Solution==<br /> &lt;center&gt;[[File:2008_AMC_12B_25_II.JPG‎]]&lt;/center&gt;<br /> <br /> Let &lt;math&gt;AP&lt;/math&gt; and &lt;math&gt;BQ&lt;/math&gt; meet &lt;math&gt;CD&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;, respectively.<br /> <br /> Since &lt;math&gt;\angle APD=90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ADP=\angle XDP&lt;/math&gt;, and they share &lt;math&gt;DP&lt;/math&gt;, triangles &lt;math&gt;APD&lt;/math&gt; and &lt;math&gt;XPD&lt;/math&gt; are congruent.<br /> <br /> By the same reasoning, we also have that triangles &lt;math&gt;BQC&lt;/math&gt; and &lt;math&gt;YQC&lt;/math&gt; are congruent.<br /> <br /> Hence, we have &lt;math&gt;[ABQCDP]=[ABYX]+\frac{[ABCD]-[ABYX]}{2}=\frac{[ABCD]+[ABYX]}{2}&lt;/math&gt;.<br /> <br /> If we let the height of the trapezoid be &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;[ABQCDP]=\frac{\frac{11+19}{2}\cdot x+\frac{11+7}{2}\cdot x}{2}=12x&lt;/math&gt;.<br /> <br /> Thusly, if we find the height of the trapezoid and multiply it by 12, we will be done.<br /> <br /> Let the projections of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;CD&lt;/math&gt; be &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;B'&lt;/math&gt;, respectively.<br /> <br /> We have &lt;math&gt;DA'+CB'=19-11=8&lt;/math&gt;, &lt;math&gt;DA'=\sqrt{DA^2-AA'^2}=\sqrt{49-x^2}&lt;/math&gt;, and &lt;math&gt;CB'=\sqrt{CB^2-BB'^2}=\sqrt{25-x^2}&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\sqrt{49-x^2}+\sqrt{25-x^2}=8&lt;/math&gt;. Solving this, we easily get that &lt;math&gt;x=\frac{5\sqrt{3}}{2}&lt;/math&gt;.<br /> <br /> Multiplying this by 12, we find that the area of hexagon &lt;math&gt;ABQCDP&lt;/math&gt; is &lt;math&gt;30\sqrt{3}&lt;/math&gt;, which corresponds to answer choice &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> unitsize(0.6cm);<br /> import olympiad;<br /> pair A,B,C,D,P,Q,M,N,W,X,Y,Z;<br /> A=(11/2,5sqrt(3)/2);<br /> B=(33/2,5sqrt(3)/2);<br /> C=(19,0);<br /> D=(0,0);<br /> P=incenter(A,D,(99999,5sqrt(3)/4));<br /> Q=incenter(B,C,(-99999,5sqrt(3)/4));<br /> W=P+(0,5sqrt(3)/4);<br /> X=P-(0,5sqrt(3)/4);<br /> Y=Q+(0,5sqrt(3)/4);<br /> Z=Q-(0,5sqrt(3)/4);<br /> M=reflect(A,P)*W;<br /> N=reflect(B,Q)*Y;<br /> draw(A--B--C--D--cycle);<br /> draw(A--P--D);<br /> draw(B--Q--C);<br /> label(&quot;$A$&quot;,A,dir(135));<br /> label(&quot;$B$&quot;,B,dir(45));<br /> label(&quot;$C$&quot;,C,dir(315));<br /> label(&quot;$D$&quot;,D,dir(225));<br /> dot(&quot;$P$&quot;,P,dir(0));<br /> dot(&quot;$Q$&quot;,Q,dir(180));<br /> draw(W--X);<br /> draw(Y--Z);<br /> draw(M--P);<br /> draw(N--Q);<br /> label(&quot;$11$&quot;,midpoint(A--B),dir(90));<br /> label(&quot;$5$&quot;,midpoint(B--C),dir(45));<br /> label(&quot;$19$&quot;,midpoint(C--D),dir(270));<br /> label(&quot;$7$&quot;,midpoint(D--A),dir(135));<br /> label(&quot;$x$&quot;,midpoint(P--W),dir(0));<br /> label(&quot;$x$&quot;,midpoint(P--X),dir(0));<br /> label(&quot;$x$&quot;,midpoint(P--M),dir(225));<br /> label(&quot;$x$&quot;,midpoint(Q--Y),dir(180));<br /> label(&quot;$x$&quot;,midpoint(Q--Z),dir(180));<br /> label(&quot;$x$&quot;,midpoint(Q--N),dir(315));<br /> draw(rightanglemark(P,W,B,12.5));<br /> draw(rightanglemark(P,X,C,12.5));<br /> draw(rightanglemark(P,M,D,12.5));<br /> draw(rightanglemark(Q,Y,A,12.5));<br /> draw(rightanglemark(Q,Z,D,12.5));<br /> draw(rightanglemark(Q,N,C,12.5));<br /> &lt;/asy&gt;<br /> <br /> Since point &lt;math&gt;P&lt;/math&gt; is the intersection of the angle bisectors of &lt;math&gt;\angle A&lt;/math&gt; and &lt;math&gt;\angle D&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt; is equidistant from &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{AD}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt;. Likewise, point &lt;math&gt;Q&lt;/math&gt; is equidistant from &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt;. Because both points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are equidistant from &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; and the distance between &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; is constant, the common distances from each of the points to the mentioned segments is equal for &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Call this distance &lt;math&gt;x&lt;/math&gt;.<br /> <br /> The distance between a point and a line is the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;\overline{AD}&lt;/math&gt; is &lt;math&gt;x&lt;/math&gt;, so the area of &lt;math&gt;\triangle ADP&lt;/math&gt; is equal to &lt;math&gt;\frac12\cdot AD\cdot x=\frac72x&lt;/math&gt;. Similarly, the area of &lt;math&gt;\triangle BCQ&lt;/math&gt; is &lt;math&gt;\frac12\cdot BC\cdot x=\frac52x&lt;/math&gt;. The altitude of the trapezoid is &lt;math&gt;2x&lt;/math&gt;, because it is the sum of the distances from either &lt;math&gt;P&lt;/math&gt; or &lt;math&gt;Q&lt;/math&gt; to &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;. This means the area of trapezoid &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\frac12(AB+CD)\cdot2x=\frac12(11+19)\cdot2x=30x&lt;/math&gt;. Now, the area of hexagon &lt;math&gt;ABQCDP&lt;/math&gt; is the area of trapezoid &lt;math&gt;ABCD&lt;/math&gt;, minus the areas of triangles &lt;math&gt;ADP&lt;/math&gt; and &lt;math&gt;BCQ&lt;/math&gt;. This is &lt;math&gt;30x-\frac72x-\frac52x=24x&lt;/math&gt;. Now it remains to find &lt;math&gt;x&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(0.6cm);<br /> import olympiad;<br /> pair A,B,C,D,R,S;<br /> A=(11/2,5sqrt(3)/2);<br /> B=(33/2,5sqrt(3)/2);<br /> C=(19,0);<br /> D=(0,0);<br /> R=(11/2,0);<br /> S=(33/2,0);<br /> draw(A--B--C--D--cycle);<br /> draw(A--R);<br /> draw(B--S);<br /> label(&quot;$A$&quot;,A,dir(135));<br /> label(&quot;$B$&quot;,B,dir(45));<br /> label(&quot;$C$&quot;,C,dir(315));<br /> label(&quot;$D$&quot;,D,dir(225));<br /> label(&quot;$R$&quot;,R,dir(270));<br /> label(&quot;$S$&quot;,S,dir(270));<br /> label(&quot;$11$&quot;,midpoint(A--B),dir(90));<br /> label(&quot;$5$&quot;,midpoint(B--C),dir(45));<br /> label(&quot;$11$&quot;,midpoint(R--S),dir(270));<br /> label(&quot;$7$&quot;,midpoint(D--A),dir(135));<br /> label(&quot;$r$&quot;,midpoint(R--D),dir(270));<br /> label(&quot;$s$&quot;,midpoint(C--S),dir(270));<br /> label(&quot;$19$&quot;,midpoint(C--D),5*dir(270));<br /> label(&quot;$2x$&quot;,midpoint(A--R),dir(0));<br /> label(&quot;$2x$&quot;,midpoint(B--S),dir(180));<br /> draw(rightanglemark(A,R,D,15));<br /> draw(rightanglemark(B,S,C,15));<br /> &lt;/asy&gt;<br /> <br /> We let &lt;math&gt;R&lt;/math&gt; and &lt;math&gt;S&lt;/math&gt; be the feet of the altitudes of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, respectively, to &lt;math&gt;\overline{CD}&lt;/math&gt;. We define &lt;math&gt;r=RD&lt;/math&gt; and &lt;math&gt;s=SC&lt;/math&gt;. We know that &lt;math&gt;AB=RS&lt;/math&gt;, so &lt;math&gt;RS=11&lt;/math&gt; and &lt;math&gt;r+s=19-11=8&lt;/math&gt;. By the Pythagorean Theorem on &lt;math&gt;\triangle ADR&lt;/math&gt; and &lt;math&gt;\triangle BCS&lt;/math&gt;, we get &lt;math&gt;r^2+(2x)^2=7^2&lt;/math&gt; and &lt;math&gt;s^2+(2x)^2=5^2&lt;/math&gt;, respectively. Subtracting the second equation from the first gives us &lt;math&gt;r^2-s^2=49-25=24&lt;/math&gt;. The left hand side of this equation is a difference of squares and factors to &lt;math&gt;(r+s)(r-s)&lt;/math&gt;. We know that &lt;math&gt;r+s=8&lt;/math&gt;, so &lt;math&gt;r-s=\frac{24}8=3&lt;/math&gt;. Now we can solve for &lt;math&gt;r&lt;/math&gt; by adding the two equations we just got to see that &lt;math&gt;2r=11&lt;/math&gt;, or &lt;math&gt;r=\frac{11}2&lt;/math&gt;.<br /> <br /> We now solve for &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;r^2+(2x)^2=49&lt;/math&gt;, so &lt;math&gt;(2x)^2=49-\left(\frac{11}2\right)^2=\frac{75}4&lt;/math&gt; and &lt;math&gt;2x=\frac{5\sqrt3}2&lt;/math&gt;. We multiply both sides of this equation by &lt;math&gt;12&lt;/math&gt; to get &lt;math&gt;24x=30\sqrt3&lt;/math&gt;. However, the area of hexagon &lt;math&gt;ABQCDP&lt;/math&gt; is &lt;math&gt;24x&lt;/math&gt;, so the answer is &lt;math&gt;30\sqrt 3&lt;/math&gt;, or answer choice &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_13&diff=93057 2018 AIME I Problems/Problem 13 2018-03-09T02:57:22Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; have side lengths &lt;math&gt;AB=30&lt;/math&gt;, &lt;math&gt;BC=32&lt;/math&gt;, and &lt;math&gt;AC=34&lt;/math&gt;. Point &lt;math&gt;X&lt;/math&gt; lies in the interior of &lt;math&gt;\overline{BC}&lt;/math&gt;, and points &lt;math&gt;I_1&lt;/math&gt; and &lt;math&gt;I_2&lt;/math&gt; are the incenters of &lt;math&gt;\triangle ABX&lt;/math&gt; and &lt;math&gt;\triangle ACX&lt;/math&gt;, respectively. Find the minimum possible area of &lt;math&gt;\triangle AI_1I_2&lt;/math&gt; as &lt;math&gt;X&lt;/math&gt; varies along &lt;math&gt;\overline{BC}&lt;/math&gt;.<br /> <br /> ==Solution==<br /> First note that &lt;cmath&gt;\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2&lt;/cmath&gt; is a constant not depending on &lt;math&gt;X&lt;/math&gt;, so by &lt;math&gt;[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2&lt;/math&gt; it suffices to minimize &lt;math&gt;(AI_1)(AI_2)&lt;/math&gt;. Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = AC&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt;, and &lt;math&gt;\alpha = \angle AXB&lt;/math&gt;. Remark that &lt;cmath&gt;\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.&lt;/cmath&gt; Applying the Law of Sines to &lt;math&gt;\triangle ABI_1&lt;/math&gt; gives &lt;cmath&gt;\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.&lt;/cmath&gt; Analogously one can derive &lt;math&gt;AI_2 = \tfrac{b\sin\frac C2}{\sin\frac\alpha 2}&lt;/math&gt;, and so &lt;cmath&gt;[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,&lt;/cmath&gt; with equality when &lt;math&gt;\alpha = 90^\circ&lt;/math&gt;, that is, when &lt;math&gt;X&lt;/math&gt; is the foot of the perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\overline{BC}&lt;/math&gt;. In this case the desired area is &lt;math&gt;bc\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2&lt;/math&gt;. To make this feasible to compute, note that &lt;cmath&gt;\sin\frac A2=\sqrt{\frac{1+\cos A}2}=\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.&lt;/cmath&gt; Applying similar logic to &lt;math&gt;\sin \tfrac B2&lt;/math&gt; and &lt;math&gt;\sin\tfrac C2&lt;/math&gt; and simplifying yields a final answer of &lt;cmath&gt;\begin{align*}bc\sin\frac A2\sin\frac B2\sin\frac C2&amp;=bc\cdot\dfrac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\&amp;=\dfrac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126}.\end{align*}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_12&diff=93056 2018 AIME I Problems/Problem 12 2018-03-09T02:54:45Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> For every subset &lt;math&gt;T&lt;/math&gt; of &lt;math&gt;U = \{ 1,2,3,\ldots,18 \}&lt;/math&gt;, let &lt;math&gt;s(T)&lt;/math&gt; be the sum of the elements of &lt;math&gt;T&lt;/math&gt;, with &lt;math&gt;s(\emptyset)&lt;/math&gt; defined to be &lt;math&gt;0&lt;/math&gt;. If &lt;math&gt;T&lt;/math&gt; is chosen at random among all subsets of &lt;math&gt;U&lt;/math&gt;, the probability that &lt;math&gt;s(T)&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Rewrite the set after mod3<br /> 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0<br /> All 0s can be omitted <br /> Case 1<br /> No 1 No 2<br /> 1<br /> Case 2<br /> 222<br /> 20<br /> Case 3<br /> 222222<br /> 1<br /> Case 4<br /> 12<br /> 6*6=36<br /> Case 5<br /> 12222<br /> 6*15=90<br /> Case 6<br /> 1122<br /> 15*15=225<br /> Case 7<br /> 1122222<br /> 15*6=90<br /> Case 8<br /> 111<br /> 20<br /> Case 9<br /> 111222<br /> 20*20=400<br /> Case 10<br /> 111222222<br /> 20<br /> Case 11<br /> 11112<br /> 15*6=90<br /> Case 12<br /> 11112222<br /> 15*15=225<br /> Case 13<br /> 1111122<br /> 6*15=90<br /> Case 14<br /> 1111122222<br /> 6*6=36<br /> Case 15<br /> 111111<br /> 1<br /> Case 16<br /> 111111222<br /> 20<br /> Case 17<br /> 111111222222<br /> 1<br /> Total 1+20+1+36+90+225+90+20+400+20+90+225+90+36+1+20+1=484+360+450+72=1366<br /> P=1362/2^12=683/2^11<br /> ANS=683<br /> <br /> <br /> ==Solution 2==<br /> Consider the numbers {1,4,7,10,13,16}. Each of those are congruent to 1mod3. There is &lt;math&gt;{6 \choose 0}=1&lt;/math&gt; way to choose zero numbers &lt;math&gt;{6 \choose 1}=6&lt;/math&gt; ways to choose 1 and so on. There ends up being &lt;math&gt;{6 \choose 0}+{6 \choose 3}+{6 \choose 6}&lt;/math&gt; possible subsets congruent to 0mod 3. There are &lt;math&gt;2^6=64&lt;/math&gt; possible subsets of these numbers.<br /> <br /> ==Solution 3==<br /> Notice that six numbers are &lt;math&gt;0\pmod3&lt;/math&gt;, six are &lt;math&gt;1\pmod3&lt;/math&gt;, and six are &lt;math&gt;2\pmod3&lt;/math&gt;. Having numbers &lt;math&gt;0\pmod3&lt;/math&gt; will not change the remainder when &lt;math&gt;s(T)&lt;/math&gt; is divided by &lt;math&gt;3&lt;/math&gt;, so we can choose any number of these in our subset. We ignore these for now. The number of numbers that are &lt;math&gt;1\pmod3&lt;/math&gt;, minus the number of numbers that are &lt;math&gt;2\pmod3&lt;/math&gt;, must be a multiple of &lt;math&gt;3&lt;/math&gt;, possibly zero or negative. We can now split into cases based on how many numbers that are &lt;math&gt;1\pmod3&lt;/math&gt; are in the set.<br /> <br /> Case 1- &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, or &lt;math&gt;6&lt;/math&gt; integers: There can be &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, or &lt;math&gt;6&lt;/math&gt; integers that are &lt;math&gt;2\pmod3&lt;/math&gt;. We can choose these in &lt;math&gt;\left(\binom60+\binom63+\binom66\right)\cdot\left(\binom60+\binom63+\binom66\right)=(1+20+1)^2=484&lt;/math&gt; ways.<br /> <br /> Case 2- &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; integers: There can be &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; integers that are &lt;math&gt;2\pmod3&lt;/math&gt;. We can choose these in &lt;math&gt;\left(\binom61+\binom64\right)\cdot\left(\binom62+\binom65\right)=(6+15)^2=441&lt;/math&gt; ways.<br /> <br /> Case 3- &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; integers: There can be &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; integers that are &lt;math&gt;2\pmod3&lt;/math&gt;. We can choose these in &lt;math&gt;\left(\binom62+\binom65\right)\cdot\left(\binom61+\binom64\right)=(15+6)^2=441&lt;/math&gt; ways.<br /> <br /> Adding these up, we get that there are &lt;math&gt;1366&lt;/math&gt; ways to choose the numbers such that their sum is a multiple of three. Putting back in the possibility that there can be multiples of &lt;math&gt;3&lt;/math&gt; in our set, we have that there are &lt;math&gt;1366\cdot\left(\binom60+\binom61+\binom62+\binom63+\binom64+\binom65+\binom66+\right)=1366\cdot2^6&lt;/math&gt; subsets &lt;math&gt;T&lt;/math&gt; with a sum that is a multiple of &lt;math&gt;3&lt;/math&gt;. Since there are &lt;math&gt;2^{18}&lt;/math&gt; total subsets, the probability is &lt;math&gt;\frac{1366\cdot2^6}{2^{18}}=\frac{683}{2^{11}}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{683}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_12&diff=93055 2018 AIME I Problems/Problem 12 2018-03-09T02:51:59Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> For every subset &lt;math&gt;T&lt;/math&gt; of &lt;math&gt;U = \{ 1,2,3,\ldots,18 \}&lt;/math&gt;, let &lt;math&gt;s(T)&lt;/math&gt; be the sum of the elements of &lt;math&gt;T&lt;/math&gt;, with &lt;math&gt;s(\emptyset)&lt;/math&gt; defined to be &lt;math&gt;0&lt;/math&gt;. If &lt;math&gt;T&lt;/math&gt; is chosen at random among all subsets of &lt;math&gt;U&lt;/math&gt;, the probability that &lt;math&gt;s(T)&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Rewrite the set after mod3<br /> 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0<br /> All 0s can be omitted <br /> Case 1<br /> No 1 No 2<br /> 1<br /> Case 2<br /> 222<br /> 20<br /> Case 3<br /> 222222<br /> 1<br /> Case 4<br /> 12<br /> 6*6=36<br /> Case 5<br /> 12222<br /> 6*15=90<br /> Case 6<br /> 1122<br /> 15*15=225<br /> Case 7<br /> 1122222<br /> 15*6=90<br /> Case 8<br /> 111<br /> 20<br /> Case 9<br /> 111222<br /> 20*20=400<br /> Case 10<br /> 111222222<br /> 20<br /> Case 11<br /> 11112<br /> 15*6=90<br /> Case 12<br /> 11112222<br /> 15*15=225<br /> Case 13<br /> 1111122<br /> 6*15=90<br /> Case 14<br /> 1111122222<br /> 6*6=36<br /> Case 15<br /> 111111<br /> 1<br /> Case 16<br /> 111111222<br /> 20<br /> Case 17<br /> 111111222222<br /> 1<br /> Total 1+20+1+36+90+225+90+20+400+20+90+225+90+36+1+20+1=484+360+450+72=1366<br /> P=1362/2^12=683/2^11<br /> ANS=683<br /> <br /> <br /> ==Solution 2==<br /> Consider the numbers {1,4,7,10,13,16}. Each of those are congruent to 1mod3. There is &lt;math&gt;{6 \choose 0}=1&lt;/math&gt; way to choose zero numbers &lt;math&gt;{6 \choose 1}=6&lt;/math&gt; ways to choose 1 and so on. There ends up being &lt;math&gt;{6 \choose 0}+{6 \choose 3}+{6 \choose 6}&lt;/math&gt; possible subsets congruent to 0mod 3. There are &lt;math&gt;2^6=64&lt;/math&gt; possible subsets of these numbers.<br /> <br /> ==Solution 3==<br /> Notice that six numbers are &lt;math&gt;0\pmod3&lt;/math&gt;, six are &lt;math&gt;1\pmod3&lt;/math&gt;, and six are &lt;math&gt;2\pmod3&lt;/math&gt;. Having numbers &lt;math&gt;0\pmod3&lt;/math&gt; will not change the remainder when &lt;math&gt;s(T)&lt;/math&gt; is divided by &lt;math&gt;3&lt;/math&gt;, so we can choose any number of these in our subset. We ignore these for now. The number of numbers that are &lt;math&gt;1\pmod3&lt;/math&gt;, minus the number of numbers that are &lt;math&gt;2\pmod3&lt;/math&gt;, must be a multiple of &lt;math&gt;3&lt;/math&gt;, possibly zero or negative. We can now split into cases based on how many numbers that are &lt;math&gt;1\pmod3&lt;/math&gt; are in the set.<br /> <br /> Case 1- &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, or &lt;math&gt;6&lt;/math&gt; integers: There can be &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, or &lt;math&gt;6&lt;/math&gt; integers that are &lt;math&gt;2\pmod3&lt;/math&gt;. We can choose these in &lt;math&gt;\left(\binom60+\binom63+\binom66\right)\cdot\left(\binom60+\binom63+\binom66\right)=(1+20+1)^2=484&lt;/math&gt; ways.<br /> <br /> Case 2- &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; integers: There can be &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; integers that are &lt;math&gt;2\pmod3&lt;/math&gt;. We can choose these in &lt;math&gt;\left(\binom61+\binom64\right)\cdot\left(\binom62+\binom65\right)=(6+15)^2=441&lt;/math&gt; ways.<br /> <br /> Case 3- &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; integers: There can be &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; integers that are &lt;math&gt;2\pmod3&lt;/math&gt;. We can choose these in &lt;math&gt;\left(\binom62+\binom65\right)\cdot\left(\binom61+\binom64\right)=(15+6)^2=441&lt;/math&gt; ways.<br /> <br /> Adding these up, we get that there are &lt;math&gt;1366&lt;/math&gt; ways to choose the numbers such that their sum is a multiple of three. Putting back in the possibility that there can be multiples of &lt;math&gt;3&lt;/math&gt; in our set, we have that there are &lt;math&gt;1366\cdot\left(\binom60+\binom61+\binom62+\binom63+\binom64+\binom65+\binom66+\right)=1366\cdot2^6&lt;/math&gt; subsets &lt;math&gt;T&lt;/math&gt; with a sum that is a multiple of &lt;math&gt;3&lt;/math&gt;. Since there are &lt;math&gt;2^{18}&lt;/math&gt; total subsets, the probability is &lt;math&gt;\frac{1366\cdot2^6}{2^{18}}=\frac{683}{2^{11}}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{683}&lt;/math&gt;.</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=Aops_font&diff=88751 Aops font 2017-12-03T00:52:06Z <p>Math101010: /* Full List */</p> <hr /> <div>The Aops font is a font that can be used to make certain special symbols in the forum. You enclose some text inside [aops] and [/aops] and the text is rendered in the aops font.<br /> <br /> = Purpose=<br /> The [aops][/aops] tags allow you to render items from the custom font that AoPS uses -- you'll recognize these icons from the rest of the site.<br /> <br /> For example, to type the symbol for the re-sort button &lt;span class=&quot;aops-font&quot;&gt;|&lt;/span&gt;, you would type <br /> :[aops]|[/aops]<br /> <br /> =Use=<br /> == In the Forums==<br /> To use the font in the forums, simply put [aops] before your text and [/aops] after it. The text will be converted as described in the tables below.<br /> == In the Wiki==<br /> Because bbcode does not work in the AoPS wiki, you need to use the following HTML code:<br /> :&lt;code&gt; &lt;nowiki&gt;&lt;span class=&quot;aops-font&quot;&gt;text here!&lt;/span&gt;&lt;/nowiki&gt;&lt;/code&gt;<br /> <br /> =Characters=<br /> ==Commonly Used==<br /> &lt;div style=&quot;font-size:large;&quot;&gt;<br /> {| class=&quot;wikitable&quot;<br /> ! Symbol<br /> ! What to type<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;Y&lt;/span&gt;<br /> | Y<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;a&lt;/span&gt;<br /> | a<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;f&lt;/span&gt;<br /> | f<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;h&lt;/span&gt;<br /> | h<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;j&lt;/span&gt;<br /> | j<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;3&lt;/span&gt;<br /> | 3<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;i&lt;/span&gt;<br /> | i<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;X&lt;/span&gt;<br /> | X<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;9&lt;/span&gt;<br /> | 9<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;k&lt;/span&gt;<br /> | k<br /> |}<br /> &lt;/div&gt;<br /> <br /> ==Full List==<br /> &lt;div style=&quot;font-size:large;&quot;&gt;<br /> {| class=&quot;wikitable&quot;<br /> ! Character<br /> ! Aops font<br /> ! Character<br /> ! Aops font<br /> |-<br /> | a<br /> | &lt;span class=&quot;aops-font&quot;&gt;a&lt;/span&gt;<br /> | A<br /> | &lt;span class=&quot;aops-font&quot;&gt;A&lt;/span&gt;<br /> |-<br /> | b<br /> | &lt;span class=&quot;aops-font&quot;&gt;b&lt;/span&gt;<br /> | B<br /> | &lt;span class=&quot;aops-font&quot;&gt;B&lt;/span&gt;<br /> |-<br /> | c<br /> | &lt;span class=&quot;aops-font&quot;&gt;c&lt;/span&gt;<br /> | C<br /> | &lt;span class=&quot;aops-font&quot;&gt;C&lt;/span&gt;<br /> |-<br /> | d<br /> | &lt;span class=&quot;aops-font&quot;&gt;d&lt;/span&gt;<br /> | D<br /> | &lt;span class=&quot;aops-font&quot;&gt;D&lt;/span&gt;<br /> |-<br /> | e<br /> | &lt;span class=&quot;aops-font&quot;&gt;e&lt;/span&gt;<br /> | E<br /> | &lt;span class=&quot;aops-font&quot;&gt;E&lt;/span&gt;<br /> |-<br /> | f<br /> | &lt;span class=&quot;aops-font&quot;&gt;f&lt;/span&gt;<br /> | F<br /> | &lt;span class=&quot;aops-font&quot;&gt;F&lt;/span&gt;<br /> |-<br /> | g<br /> | &lt;span class=&quot;aops-font&quot;&gt;g&lt;/span&gt;<br /> | G<br /> | &lt;span class=&quot;aops-font&quot;&gt;G&lt;/span&gt;<br /> |-<br /> | h<br /> | &lt;span class=&quot;aops-font&quot;&gt;h&lt;/span&gt;<br /> | H<br /> | &lt;span class=&quot;aops-font&quot;&gt;H&lt;/span&gt;<br /> |-<br /> | i<br /> | &lt;span class=&quot;aops-font&quot;&gt;i&lt;/span&gt;<br /> | I<br /> | &lt;span class=&quot;aops-font&quot;&gt;I&lt;/span&gt;<br /> |-<br /> | j<br /> | &lt;span class=&quot;aops-font&quot;&gt;j&lt;/span&gt;<br /> | J<br /> | &lt;span class=&quot;aops-font&quot;&gt;J&lt;/span&gt;<br /> |-<br /> | k<br /> | &lt;span class=&quot;aops-font&quot;&gt;k&lt;/span&gt;<br /> | K<br /> | &lt;span class=&quot;aops-font&quot;&gt;K&lt;/span&gt;<br /> |-<br /> | l<br /> | &lt;span class=&quot;aops-font&quot;&gt;l&lt;/span&gt;<br /> | L<br /> | &lt;span class=&quot;aops-font&quot;&gt;L&lt;/span&gt;<br /> |-<br /> | m<br /> | &lt;span class=&quot;aops-font&quot;&gt;m&lt;/span&gt;<br /> | M<br /> | &lt;span class=&quot;aops-font&quot;&gt;M&lt;/span&gt;<br /> |-<br /> | n<br /> | &lt;span class=&quot;aops-font&quot;&gt;n&lt;/span&gt;<br /> | N<br /> | &lt;span class=&quot;aops-font&quot;&gt;N&lt;/span&gt;<br /> |-<br /> | o<br /> | &lt;span class=&quot;aops-font&quot;&gt;o&lt;/span&gt;<br /> | O<br /> | &lt;span class=&quot;aops-font&quot;&gt;O&lt;/span&gt;<br /> |-<br /> | p<br /> | &lt;span class=&quot;aops-font&quot;&gt;p&lt;/span&gt;<br /> | P<br /> | &lt;span class=&quot;aops-font&quot;&gt;P&lt;/span&gt;<br /> |-<br /> | q<br /> | &lt;span class=&quot;aops-font&quot;&gt;q&lt;/span&gt;<br /> | Q<br /> | &lt;span class=&quot;aops-font&quot;&gt;Q&lt;/span&gt;<br /> |-<br /> | r<br /> | &lt;span class=&quot;aops-font&quot;&gt;r&lt;/span&gt;<br /> | R<br /> | &lt;span class=&quot;aops-font&quot;&gt;R&lt;/span&gt;<br /> |-<br /> | s<br /> | &lt;span class=&quot;aops-font&quot;&gt;s&lt;/span&gt;<br /> | S<br /> | &lt;span class=&quot;aops-font&quot;&gt;S&lt;/span&gt;<br /> |-<br /> | t<br /> | &lt;span class=&quot;aops-font&quot;&gt;t&lt;/span&gt;<br /> | T<br /> | &lt;span class=&quot;aops-font&quot;&gt;T&lt;/span&gt;<br /> |-<br /> | u<br /> | &lt;span class=&quot;aops-font&quot;&gt;u&lt;/span&gt;<br /> | U<br /> | &lt;span class=&quot;aops-font&quot;&gt;U&lt;/span&gt;<br /> |-<br /> | v<br /> | &lt;span class=&quot;aops-font&quot;&gt;v&lt;/span&gt;<br /> | V<br /> | &lt;span class=&quot;aops-font&quot;&gt;V&lt;/span&gt;<br /> |-<br /> | w<br /> | &lt;span class=&quot;aops-font&quot;&gt;w&lt;/span&gt;<br /> | W<br /> | &lt;span class=&quot;aops-font&quot;&gt;W&lt;/span&gt;<br /> |-<br /> | x<br /> | &lt;span class=&quot;aops-font&quot;&gt;x&lt;/span&gt;<br /> | X<br /> | &lt;span class=&quot;aops-font&quot;&gt;X&lt;/span&gt;<br /> |-<br /> | y<br /> | &lt;span class=&quot;aops-font&quot;&gt;y&lt;/span&gt;<br /> | Y<br /> | &lt;span class=&quot;aops-font&quot;&gt;Y&lt;/span&gt;<br /> |-<br /> | z<br /> | &lt;span class=&quot;aops-font&quot;&gt;z&lt;/span&gt;<br /> | Z<br /> | &lt;span class=&quot;aops-font&quot;&gt;Z&lt;/span&gt;<br /> |-<br /> | 1<br /> | &lt;span class=&quot;aops-font&quot;&gt;1&lt;/span&gt;<br /> | !<br /> | &lt;span class=&quot;aops-font&quot;&gt;!&lt;/span&gt;<br /> |-<br /> | 2<br /> | &lt;span class=&quot;aops-font&quot;&gt;2&lt;/span&gt;<br /> | @<br /> | &lt;span class=&quot;aops-font&quot;&gt;@&lt;/span&gt;<br /> |-<br /> | 3<br /> | &lt;span class=&quot;aops-font&quot;&gt;3&lt;/span&gt;<br /> | #<br /> | &lt;span class=&quot;aops-font&quot;&gt;#&lt;/span&gt;<br /> |-<br /> | 4<br /> | &lt;span class=&quot;aops-font&quot;&gt;4&lt;/span&gt;<br /> | &amp;#36;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;#36;&lt;/span&gt;<br /> |-<br /> | 5<br /> | &lt;span class=&quot;aops-font&quot;&gt;5&lt;/span&gt;<br /> | %<br /> | &lt;span class=&quot;aops-font&quot;&gt;%&lt;/span&gt;<br /> |-<br /> | 6<br /> | &lt;span class=&quot;aops-font&quot;&gt;6&lt;/span&gt;<br /> | ^<br /> | &lt;span class=&quot;aops-font&quot;&gt;^&lt;/span&gt;<br /> |-<br /> | 7<br /> | &lt;span class=&quot;aops-font&quot;&gt;7&lt;/span&gt;<br /> | &amp;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;&lt;/span&gt;<br /> |-<br /> | 8<br /> | &lt;span class=&quot;aops-font&quot;&gt;8&lt;/span&gt;<br /> | *<br /> | &lt;span class=&quot;aops-font&quot;&gt;*&lt;/span&gt;<br /> |-<br /> | 9<br /> | &lt;span class=&quot;aops-font&quot;&gt;9&lt;/span&gt;<br /> | (<br /> | &lt;span class=&quot;aops-font&quot;&gt;(&lt;/span&gt;<br /> |-<br /> | 0<br /> | &lt;span class=&quot;aops-font&quot;&gt;0&lt;/span&gt;<br /> | )<br /> | &lt;span class=&quot;aops-font&quot;&gt;)&lt;/span&gt;<br /> |-<br /> | -<br /> | &lt;span class=&quot;aops-font&quot;&gt;-&lt;/span&gt;<br /> | _<br /> | &lt;span class=&quot;aops-font&quot;&gt;_&lt;/span&gt;<br /> |-<br /> | =<br /> | &lt;span class=&quot;aops-font&quot;&gt;=&lt;/span&gt; <br /> | +<br /> | &lt;span class=&quot;aops-font&quot;&gt;+&lt;/span&gt; <br /> |-<br /> | ,<br /> | &lt;span class=&quot;aops-font&quot;&gt;,&lt;/span&gt;<br /> | .<br /> | &lt;span class=&quot;aops-font&quot;&gt;.&lt;/span&gt;<br /> |-<br /> | /<br /> | &lt;span class=&quot;aops-font&quot;&gt;/&lt;/span&gt;<br /> | ?<br /> | &lt;span class=&quot;aops-font&quot;&gt;?&lt;/span&gt;<br /> |-<br /> | &lt;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;lt;&lt;/span&gt;<br /> | &gt;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&gt;&lt;/span&gt;<br /> |-<br /> |;<br /> | &lt;span class=&quot;aops-font&quot;&gt;;&lt;/span&gt;<br /> |:<br /> | &lt;span class=&quot;aops-font&quot;&gt;:&lt;/span&gt;<br /> |-<br /> |'<br /> | &lt;span class=&quot;aops-font&quot;&gt;'&lt;/span&gt;<br /> |&quot;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&quot;&lt;/span&gt;<br /> |-<br /> | {<br /> | &lt;span class=&quot;aops-font&quot;&gt;{&lt;/span&gt;<br /> | }<br /> | &lt;span class=&quot;aops-font&quot;&gt;}&lt;/span&gt;<br /> |-<br /> | –<br /> | &lt;span class=&quot;aops-font&quot;&gt;–&lt;/span&gt;<br /> | —<br /> | &lt;span class=&quot;aops-font&quot;&gt;—&lt;/span&gt;<br /> |-<br /> | €<br /> | &lt;span class=&quot;aops-font&quot;&gt;€&lt;/span&gt;<br /> | ·<br /> | &lt;span class=&quot;aops-font&quot;&gt;·&lt;/span&gt;<br /> |-<br /> | ∂<br /> | &lt;span class=&quot;aops-font&quot;&gt;∂&lt;/span&gt;<br /> | ∆<br /> | &lt;span class=&quot;aops-font&quot;&gt;∆&lt;/span&gt;<br /> |-<br /> | ‹<br /> | &lt;span class=&quot;aops-font&quot;&gt;‹&lt;/span&gt;<br /> | ›<br /> | &lt;span class=&quot;aops-font&quot;&gt;›&lt;/span&gt;<br /> |}<br /> &lt;/div&gt; <br /> <br /> [[Category:AoPS forums]]</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=Aops_font&diff=88750 Aops font 2017-12-03T00:42:31Z <p>Math101010: /* Full List */</p> <hr /> <div>The Aops font is a font that can be used to make certain special symbols in the forum. You enclose some text inside [aops] and [/aops] and the text is rendered in the aops font.<br /> <br /> = Purpose=<br /> The [aops][/aops] tags allow you to render items from the custom font that AoPS uses -- you'll recognize these icons from the rest of the site.<br /> <br /> For example, to type the symbol for the re-sort button &lt;span class=&quot;aops-font&quot;&gt;|&lt;/span&gt;, you would type <br /> :[aops]|[/aops]<br /> <br /> =Use=<br /> == In the Forums==<br /> To use the font in the forums, simply put [aops] before your text and [/aops] after it. The text will be converted as described in the tables below.<br /> == In the Wiki==<br /> Because bbcode does not work in the AoPS wiki, you need to use the following HTML code:<br /> :&lt;code&gt; &lt;nowiki&gt;&lt;span class=&quot;aops-font&quot;&gt;text here!&lt;/span&gt;&lt;/nowiki&gt;&lt;/code&gt;<br /> <br /> =Characters=<br /> ==Commonly Used==<br /> &lt;div style=&quot;font-size:large;&quot;&gt;<br /> {| class=&quot;wikitable&quot;<br /> ! Symbol<br /> ! What to type<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;Y&lt;/span&gt;<br /> | Y<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;a&lt;/span&gt;<br /> | a<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;f&lt;/span&gt;<br /> | f<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;h&lt;/span&gt;<br /> | h<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;j&lt;/span&gt;<br /> | j<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;3&lt;/span&gt;<br /> | 3<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;i&lt;/span&gt;<br /> | i<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;X&lt;/span&gt;<br /> | X<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;9&lt;/span&gt;<br /> | 9<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;k&lt;/span&gt;<br /> | k<br /> |}<br /> &lt;/div&gt;<br /> <br /> ==Full List==<br /> &lt;div style=&quot;font-size:large;&quot;&gt;<br /> {| class=&quot;wikitable&quot;<br /> ! Character<br /> ! Aops font<br /> ! Character<br /> ! Aops font<br /> |-<br /> | a<br /> | &lt;span class=&quot;aops-font&quot;&gt;a&lt;/span&gt;<br /> | A<br /> | &lt;span class=&quot;aops-font&quot;&gt;A&lt;/span&gt;<br /> |-<br /> | b<br /> | &lt;span class=&quot;aops-font&quot;&gt;b&lt;/span&gt;<br /> | B<br /> | &lt;span class=&quot;aops-font&quot;&gt;B&lt;/span&gt;<br /> |-<br /> | c<br /> | &lt;span class=&quot;aops-font&quot;&gt;c&lt;/span&gt;<br /> | C<br /> | &lt;span class=&quot;aops-font&quot;&gt;C&lt;/span&gt;<br /> |-<br /> | d<br /> | &lt;span class=&quot;aops-font&quot;&gt;d&lt;/span&gt;<br /> | D<br /> | &lt;span class=&quot;aops-font&quot;&gt;D&lt;/span&gt;<br /> |-<br /> | e<br /> | &lt;span class=&quot;aops-font&quot;&gt;e&lt;/span&gt;<br /> | E<br /> | &lt;span class=&quot;aops-font&quot;&gt;E&lt;/span&gt;<br /> |-<br /> | f<br /> | &lt;span class=&quot;aops-font&quot;&gt;f&lt;/span&gt;<br /> | F<br /> | &lt;span class=&quot;aops-font&quot;&gt;F&lt;/span&gt;<br /> |-<br /> | g<br /> | &lt;span class=&quot;aops-font&quot;&gt;g&lt;/span&gt;<br /> | G<br /> | &lt;span class=&quot;aops-font&quot;&gt;G&lt;/span&gt;<br /> |-<br /> | h<br /> | &lt;span class=&quot;aops-font&quot;&gt;h&lt;/span&gt;<br /> | H<br /> | &lt;span class=&quot;aops-font&quot;&gt;H&lt;/span&gt;<br /> |-<br /> | i<br /> | &lt;span class=&quot;aops-font&quot;&gt;i&lt;/span&gt;<br /> | I<br /> | &lt;span class=&quot;aops-font&quot;&gt;I&lt;/span&gt;<br /> |-<br /> | j<br /> | &lt;span class=&quot;aops-font&quot;&gt;j&lt;/span&gt;<br /> | J<br /> | &lt;span class=&quot;aops-font&quot;&gt;J&lt;/span&gt;<br /> |-<br /> | k<br /> | &lt;span class=&quot;aops-font&quot;&gt;k&lt;/span&gt;<br /> | K<br /> | &lt;span class=&quot;aops-font&quot;&gt;K&lt;/span&gt;<br /> |-<br /> | l<br /> | &lt;span class=&quot;aops-font&quot;&gt;l&lt;/span&gt;<br /> | L<br /> | &lt;span class=&quot;aops-font&quot;&gt;L&lt;/span&gt;<br /> |-<br /> | m<br /> | &lt;span class=&quot;aops-font&quot;&gt;m&lt;/span&gt;<br /> | M<br /> | &lt;span class=&quot;aops-font&quot;&gt;M&lt;/span&gt;<br /> |-<br /> | n<br /> | &lt;span class=&quot;aops-font&quot;&gt;n&lt;/span&gt;<br /> | N<br /> | &lt;span class=&quot;aops-font&quot;&gt;N&lt;/span&gt;<br /> |-<br /> | o<br /> | &lt;span class=&quot;aops-font&quot;&gt;o&lt;/span&gt;<br /> | O<br /> | &lt;span class=&quot;aops-font&quot;&gt;O&lt;/span&gt;<br /> |-<br /> | p<br /> | &lt;span class=&quot;aops-font&quot;&gt;p&lt;/span&gt;<br /> | P<br /> | &lt;span class=&quot;aops-font&quot;&gt;P&lt;/span&gt;<br /> |-<br /> | q<br /> | &lt;span class=&quot;aops-font&quot;&gt;q&lt;/span&gt;<br /> | Q<br /> | &lt;span class=&quot;aops-font&quot;&gt;Q&lt;/span&gt;<br /> |-<br /> | r<br /> | &lt;span class=&quot;aops-font&quot;&gt;r&lt;/span&gt;<br /> | R<br /> | &lt;span class=&quot;aops-font&quot;&gt;R&lt;/span&gt;<br /> |-<br /> | s<br /> | &lt;span class=&quot;aops-font&quot;&gt;s&lt;/span&gt;<br /> | S<br /> | &lt;span class=&quot;aops-font&quot;&gt;S&lt;/span&gt;<br /> |-<br /> | t<br /> | &lt;span class=&quot;aops-font&quot;&gt;t&lt;/span&gt;<br /> | T<br /> | &lt;span class=&quot;aops-font&quot;&gt;T&lt;/span&gt;<br /> |-<br /> | u<br /> | &lt;span class=&quot;aops-font&quot;&gt;u&lt;/span&gt;<br /> | U<br /> | &lt;span class=&quot;aops-font&quot;&gt;U&lt;/span&gt;<br /> |-<br /> | v<br /> | &lt;span class=&quot;aops-font&quot;&gt;v&lt;/span&gt;<br /> | V<br /> | &lt;span class=&quot;aops-font&quot;&gt;V&lt;/span&gt;<br /> |-<br /> | w<br /> | &lt;span class=&quot;aops-font&quot;&gt;w&lt;/span&gt;<br /> | W<br /> | &lt;span class=&quot;aops-font&quot;&gt;W&lt;/span&gt;<br /> |-<br /> | x<br /> | &lt;span class=&quot;aops-font&quot;&gt;x&lt;/span&gt;<br /> | X<br /> | &lt;span class=&quot;aops-font&quot;&gt;X&lt;/span&gt;<br /> |-<br /> | y<br /> | &lt;span class=&quot;aops-font&quot;&gt;y&lt;/span&gt;<br /> | Y<br /> | &lt;span class=&quot;aops-font&quot;&gt;Y&lt;/span&gt;<br /> |-<br /> | z<br /> | &lt;span class=&quot;aops-font&quot;&gt;z&lt;/span&gt;<br /> | Z<br /> | &lt;span class=&quot;aops-font&quot;&gt;Z&lt;/span&gt;<br /> |-<br /> | 1<br /> | &lt;span class=&quot;aops-font&quot;&gt;1&lt;/span&gt;<br /> | !<br /> | &lt;span class=&quot;aops-font&quot;&gt;!&lt;/span&gt;<br /> |-<br /> | 2<br /> | &lt;span class=&quot;aops-font&quot;&gt;2&lt;/span&gt;<br /> | @<br /> | &lt;span class=&quot;aops-font&quot;&gt;@&lt;/span&gt;<br /> |-<br /> | 3<br /> | &lt;span class=&quot;aops-font&quot;&gt;3&lt;/span&gt;<br /> | #<br /> | &lt;span class=&quot;aops-font&quot;&gt;#&lt;/span&gt;<br /> |-<br /> | 4<br /> | &lt;span class=&quot;aops-font&quot;&gt;4&lt;/span&gt;<br /> | &amp;#36;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;#36;&lt;/span&gt;<br /> |-<br /> | 5<br /> | &lt;span class=&quot;aops-font&quot;&gt;5&lt;/span&gt;<br /> | %<br /> | &lt;span class=&quot;aops-font&quot;&gt;%&lt;/span&gt;<br /> |-<br /> | 6<br /> | &lt;span class=&quot;aops-font&quot;&gt;6&lt;/span&gt;<br /> | ^<br /> | &lt;span class=&quot;aops-font&quot;&gt;^&lt;/span&gt;<br /> |-<br /> | 7<br /> | &lt;span class=&quot;aops-font&quot;&gt;7&lt;/span&gt;<br /> | &amp;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;&lt;/span&gt;<br /> |-<br /> | 8<br /> | &lt;span class=&quot;aops-font&quot;&gt;8&lt;/span&gt;<br /> | *<br /> | &lt;span class=&quot;aops-font&quot;&gt;*&lt;/span&gt;<br /> |-<br /> | 9<br /> | &lt;span class=&quot;aops-font&quot;&gt;9&lt;/span&gt;<br /> | (<br /> | &lt;span class=&quot;aops-font&quot;&gt;(&lt;/span&gt;<br /> |-<br /> | 0<br /> | &lt;span class=&quot;aops-font&quot;&gt;0&lt;/span&gt;<br /> | )<br /> | &lt;span class=&quot;aops-font&quot;&gt;)&lt;/span&gt;<br /> |-<br /> | -<br /> | &lt;span class=&quot;aops-font&quot;&gt;-&lt;/span&gt;<br /> | _<br /> | &lt;span class=&quot;aops-font&quot;&gt;_&lt;/span&gt;<br /> |-<br /> | =<br /> | &lt;span class=&quot;aops-font&quot;&gt;=&lt;/span&gt; <br /> | +<br /> | &lt;span class=&quot;aops-font&quot;&gt;+&lt;/span&gt; <br /> |-<br /> | ,<br /> | &lt;span class=&quot;aops-font&quot;&gt;,&lt;/span&gt;<br /> | .<br /> | &lt;span class=&quot;aops-font&quot;&gt;.&lt;/span&gt;<br /> |-<br /> | /<br /> | &lt;span class=&quot;aops-font&quot;&gt;/&lt;/span&gt;<br /> | ?<br /> | &lt;span class=&quot;aops-font&quot;&gt;?&lt;/span&gt;<br /> |-<br /> | &lt;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;lt;&lt;/span&gt;<br /> | &gt;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&gt;&lt;/span&gt;<br /> |-<br /> |;<br /> | &lt;span class=&quot;aops-font&quot;&gt;;&lt;/span&gt;<br /> |:<br /> | &lt;span class=&quot;aops-font&quot;&gt;:&lt;/span&gt;<br /> |-<br /> |'<br /> | &lt;span class=&quot;aops-font&quot;&gt;'&lt;/span&gt;<br /> |&quot;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&quot;&lt;/span&gt;<br /> |-<br /> | {<br /> | &lt;span class=&quot;aops-font&quot;&gt;{&lt;/span&gt;<br /> | }<br /> | &lt;span class=&quot;aops-font&quot;&gt;}&lt;/span&gt;<br /> |-<br /> | –<br /> | &lt;span class=&quot;aops-font&quot;&gt;–&lt;/span&gt;<br /> | —<br /> | &lt;span class=&quot;aops-font&quot;&gt;—&lt;/span&gt;<br /> |-<br /> | €<br /> | &lt;span class=&quot;aops-font&quot;&gt;€&lt;/span&gt;<br /> |}<br /> &lt;/div&gt; <br /> <br /> [[Category:AoPS forums]]</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=User:Math101010&diff=87803 User:Math101010 2017-10-14T16:07:25Z <p>Math101010: Replaced content with &quot;Why does this exist&quot;</p> <hr /> <div>Why does this exist</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=User:Math101010&diff=86416 User:Math101010 2017-07-17T03:30:21Z <p>Math101010: /* Contests */</p> <hr /> <div>Math101010 joined AoPS on March 7, 2015, shortly after the new community was released. The March 1, 2015 join date in his profile is an error. Note: The only reason that this actually has content is because Kline reported Math101010 for having a &quot;spammy&quot; user page.<br /> <br /> ==Mathematics==<br /> Math101010 enjoys doing mathematics very much. Once, he insisted on doing math instead of spending time with this family. Except he's not all that great at math. Math101010 likes assuming his own gender and referring to himself in third person.<br /> ===School System===<br /> He is taking Prealgebra.<br /> ===On AoPS===<br /> He started the classes Intro to Geometry in March, Intermediate Algebra in June, and AMC 10 problem series in October 2015. He took all intermediate classes by the end of 2016, and just finished calculus in April 2017.<br /> ===Contests===<br /> In 2015, he scored a rating of 21 in state MATHCOUNTS. He scored a 94.5 in the AMC 10 competition, where he was trolled by the MAA. ([[2015 AMC 10A Problems/Problem 14]])<br /> <br /> In 2016, he scored a 115.5 on the AMC 10A and 100.5 on the AMC 10B. He qualified for the AIME through the A contest, where he scored a 5. He still hopes to qualify for National MATHCOUNTS. He went to ARML, and got a 2 on the individual round, which was &lt;math&gt;\frac16&lt;/math&gt; of his team's total score.<br /> <br /> In 2017, he scored a 120 on the AMC 10A and a 9 on the AIME I. He refuses to talk about the AMC 10B. He completely bombed MATHCOUNTS this year. He got a 6 on the ARML individual round.<br /> <br /> ==Goals in Mathematics==<br /> He hopes to qualify for the USA(J)MO someday. For competitions, he hopes to score at least 132 on the AMC 10 and at least 10 on the AIME. He believes this can be done by taking WOOT, but his parents think it is too expensive and too hard. To date (4/20/2017), Math101010 has never solved a USA(J)MO problem without assistance, but hopes to be able to do so soon.<br /> <br /> ==Non-mathematics==<br /> Math101010 officially has no life.<br /> <br /> ==See also==<br /> [[2015 AMC 10A Problems/Problem 14]]<br /> https://www.youtube.com/watch?v=pbcEOD2rRxs<br /> https://artofproblemsolving.com/community/c116925</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems&diff=86259 2008 AMC 8 Problems 2017-07-03T15:02:28Z <p>Math101010: /* Problem 8 */</p> <hr /> <div>==Problem 1==<br /> Susan had 50 dollars to spend at the carnival. She spent 12 dollars on food and twice as much on rides. How many dollars did she have left to spend?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{12} \qquad<br /> \textbf{(B)}\ \text{14} \qquad<br /> \textbf{(C)}\ \text{26} \qquad<br /> \textbf{(D)}\ \text{38}\qquad<br /> \textbf{(E)}\ \text{50} &lt;/math&gt;<br /> <br /> <br /> [[2008 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> The ten-letter code &lt;math&gt;\text{BEST OF LUCK}&lt;/math&gt; represents the ten digits &lt;math&gt;0-9&lt;/math&gt;, in order. What 4-digit number is represented by the code word &lt;math&gt;\text{CLUE}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8671 \qquad<br /> \textbf{(B)}\ 8672 \qquad<br /> \textbf{(C)}\ 9781 \qquad<br /> \textbf{(D)}\ 9782 \qquad<br /> \textbf{(E)}\ 9872&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> If February is a month that contains Friday the &lt;math&gt;13^{\text{th}}&lt;/math&gt;, what day of the week is February 1?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Sunday} \qquad<br /> \textbf{(B)}\ \text{Monday} \qquad<br /> \textbf{(C)}\ \text{Wednesday} \qquad<br /> \textbf{(D)}\ \text{Thursday}\qquad<br /> \textbf{(E)}\ \text{Saturday} &lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> In the figure, the outer equilateral triangle has area &lt;math&gt;16&lt;/math&gt;, the inner equilateral triangle has area &lt;math&gt;1&lt;/math&gt;, and the three trapezoids are congruent. What is the area of one of the trapezoids?<br /> &lt;asy&gt;<br /> size((70));<br /> draw((0,0)--(7.5,13)--(15,0)--(0,0));<br /> draw((1.88,3.25)--(9.45,3.25));<br /> draw((11.2,0)--(7.5,6.5));<br /> draw((9.4,9.7)--(5.6,3.25));<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Barney Schwinn notices that the odometer on his bicycle reads &lt;math&gt;1441&lt;/math&gt;, a palindrome, because it reads the same forward and backward. After riding &lt;math&gt;4&lt;/math&gt; more hours that day and &lt;math&gt;6&lt;/math&gt; the next, he notices that the odometer shows another palindrome, &lt;math&gt;1661&lt;/math&gt;. What was his average speed in miles per hour?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad<br /> \textbf{(B)}\ 16\qquad<br /> \textbf{(C)}\ 18\qquad<br /> \textbf{(D)}\ 20\qquad<br /> \textbf{(E)}\ 22&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> In the figure, what is the ratio of the area of the gray squares to the area of the white squares?<br /> &lt;asy&gt;<br /> size((70));<br /> draw((10,0)--(0,10)--(-10,0)--(0,-10)--(10,0));<br /> draw((-2.5,-7.5)--(7.5,2.5));<br /> draw((-5,-5)--(5,5));<br /> draw((-7.5,-2.5)--(2.5,7.5));<br /> draw((-7.5,2.5)--(2.5,-7.5));<br /> draw((-5,5)--(5,-5));<br /> draw((-2.5,7.5)--(7.5,-2.5));<br /> fill((-10,0)--(-7.5,2.5)--(-5,0)--(-7.5,-2.5)--cycle, gray);<br /> fill((-5,0)--(0,5)--(5,0)--(0,-5)--cycle, gray);<br /> fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 &lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> If &lt;math&gt;\frac{3}{5}=\frac{M}{45}=\frac{60}{N}&lt;/math&gt;, what is &lt;math&gt;M+N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 27\qquad<br /> \textbf{(B)}\ 29 \qquad<br /> \textbf{(C)}\ 45 \qquad<br /> \textbf{(D)}\ 105\qquad<br /> \textbf{(E)}\ 127&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Candy sales from the Boosters Club from January through April are shown. What were the average sales per month in dollars?<br /> &lt;asy&gt;<br /> draw((0,0)--(36,0)--(36,24)--(0,24)--cycle);<br /> draw((0,4)--(36,4));<br /> draw((0,8)--(36,8));<br /> draw((0,12)--(36,12));<br /> draw((0,16)--(36,16));<br /> draw((0,20)--(36,20));<br /> fill((4,0)--(8,0)--(8,20)--(4,20)--cycle, black);<br /> fill((12,0)--(16,0)--(16,12)--(12,12)--cycle, black);<br /> fill((20,0)--(24,0)--(24,8)--(20,8)--cycle, black);<br /> fill((28,0)--(32,0)--(32,24)--(28,24)--cycle, black);<br /> label(&quot;120&quot;, (0,24), W);<br /> label(&quot;80&quot;, (0,16), W);<br /> label(&quot;40&quot;, (0,8), W);<br /> label(&quot;Jan&quot;, (6,0), S);<br /> label(&quot;Feb&quot;, (14,0), S);<br /> label(&quot;Mar&quot;, (22,0), S);<br /> label(&quot;Apr&quot;, (30,0), S);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ 60\qquad\textbf{(B)}\ 70\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 80\qquad\textbf{(E)}\ 85 &lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> In &lt;math&gt;2005&lt;/math&gt; Tycoon Tammy invested &lt;math&gt;100&lt;/math&gt; dollars for two years. During the the first year<br /> her investment suffered a &lt;math&gt;15\%&lt;/math&gt; loss, but during the second year the remaining<br /> investment showed a &lt;math&gt;20\%&lt;/math&gt; gain. Over the two-year period, what was the change<br /> in Tammy's investment?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 5\%\text{ loss}\qquad<br /> \textbf{(B)}\ 2\%\text{ loss}\qquad<br /> \textbf{(C)}\ 1\%\text{ gain}\qquad<br /> \textbf{(D)}\ 2\% \text{ gain} \qquad<br /> \textbf{(E)}\ 5\%\text{ gain}&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> The average age of the &lt;math&gt;6&lt;/math&gt; people in Room A is &lt;math&gt;40&lt;/math&gt;. The average age of the &lt;math&gt;4&lt;/math&gt; people in Room B is &lt;math&gt;25&lt;/math&gt;. If the two groups are combined, what is the average age of all the people?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32.5 \qquad<br /> \textbf{(B)}\ 33 \qquad<br /> \textbf{(C)}\ 33.5 \qquad<br /> \textbf{(D)}\ 34\qquad<br /> \textbf{(E)}\ 35&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Each of the &lt;math&gt;39&lt;/math&gt; students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and &lt;math&gt;26&lt;/math&gt; students have a cat. How many students have both a dog and a cat?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7\qquad<br /> \textbf{(B)}\ 13\qquad<br /> \textbf{(C)}\ 19\qquad<br /> \textbf{(D)}\ 39\qquad<br /> \textbf{(E)}\ 46&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> A ball is dropped from a height of &lt;math&gt;3&lt;/math&gt; meters. On its first bounce it rises to a height of &lt;math&gt;2&lt;/math&gt; meters. It keeps falling and bouncing to &lt;math&gt;\frac{2}{3}&lt;/math&gt; of the height it reached in the previous bounce. On which bounce will it rise to a height less than &lt;math&gt;0.5&lt;/math&gt; meters?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 4 \qquad<br /> \textbf{(C)}\ 5 \qquad<br /> \textbf{(D)}\ 6 \qquad<br /> \textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than &lt;math&gt;100&lt;/math&gt; pounds or more than &lt;math&gt;150&lt;/math&gt; pounds. So the boxes are weighed in pairs in every possible way. The results are &lt;math&gt;122&lt;/math&gt;, &lt;math&gt;125&lt;/math&gt; and &lt;math&gt;127&lt;/math&gt; pounds. What is the combined weight in pounds of the three boxes?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 160\qquad<br /> \textbf{(B)}\ 170\qquad<br /> \textbf{(C)}\ 187\qquad<br /> \textbf{(D)}\ 195\qquad<br /> \textbf{(E)}\ 354&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Three &lt;math&gt;\text{A's}&lt;/math&gt;, three &lt;math&gt;\text{B's}&lt;/math&gt;, and three &lt;math&gt;\text{C's}&lt;/math&gt; are placed in the nine spaces so that each row and column contain one of each letter. If &lt;math&gt;\text{A}&lt;/math&gt; is placed in the upper left corner, how many arrangements are possible?<br /> &lt;asy&gt;<br /> size((80));<br /> draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0));<br /> draw((3,0)--(3,9));<br /> draw((6,0)--(6,9));<br /> draw((0,3)--(9,3));<br /> draw((0,6)--(9,6));<br /> label(&quot;A&quot;, (1.5,7.5));<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 &lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> In Theresa's first &lt;math&gt;8&lt;/math&gt; basketball games, she scored &lt;math&gt;7, 4, 3, 6, 8, 3, 1&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; points. In her ninth game, she scored fewer than &lt;math&gt;10&lt;/math&gt; points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than &lt;math&gt;10&lt;/math&gt; points and her points-per-game average for the &lt;math&gt;10&lt;/math&gt; games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 35\qquad<br /> \textbf{(B)}\ 40\qquad<br /> \textbf{(C)}\ 48\qquad<br /> \textbf{(D)}\ 56\qquad<br /> \textbf{(E)}\ 72&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(1,1/2,1/4);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(1,-1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(1,0,-1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(1,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(1,0,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} \:1 : 6 \qquad\textbf{ (B)}\: 7 : 36 \qquad\textbf{(C)}\: 1 : 5 \qquad\textbf{(D)}\: 7 : 30\qquad\textbf{ (E)}\: 6 : 25&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of &lt;math&gt;50&lt;/math&gt; units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 76\qquad<br /> \textbf{(B)}\ 120\qquad<br /> \textbf{(C)}\ 128\qquad<br /> \textbf{(D)}\ 132\qquad<br /> \textbf{(E)}\ 136&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Two circles that share the same center have radii &lt;math&gt;10&lt;/math&gt; meters and &lt;math&gt;20&lt;/math&gt; meters. An aardvark runs along the path shown, starting at &lt;math&gt;A&lt;/math&gt; and ending at &lt;math&gt;K&lt;/math&gt;. How many meters does the aardvark run?<br /> &lt;asy&gt;<br /> size((150));<br /> draw((10,0)..(0,10)..(-10,0)..(0,-10)..cycle);<br /> draw((20,0)..(0,20)..(-20,0)..(0,-20)..cycle);<br /> draw((20,0)--(-20,0));<br /> draw((0,20)--(0,-20));<br /> draw((-2,21.5)..(-15.4, 15.4)..(-22,0), EndArrow);<br /> draw((-18,1)--(-12, 1), EndArrow);<br /> draw((-12,0)..(-8.3,-8.3)..(0,-12), EndArrow);<br /> draw((1,-9)--(1,9), EndArrow);<br /> draw((0,12)..(8.3, 8.3)..(12,0), EndArrow);<br /> draw((12,-1)--(18,-1), EndArrow);<br /> label(&quot;$A$&quot;, (0,20), N);<br /> label(&quot;$K$&quot;, (20,0), E);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ 10\pi+20\qquad\textbf{(B)}\ 10\pi+30\qquad\textbf{(C)}\ 10\pi+40\qquad\textbf{(D)}\ 20\pi+20\qquad \\ \textbf{(E)}\ 20\pi+40&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> Eight points are spaced around at intervals of one unit around a &lt;math&gt;2 \times 2&lt;/math&gt; square, as shown. Two of the &lt;math&gt;8&lt;/math&gt; points are chosen at random. What is the probability that the two points are one unit apart?<br /> &lt;asy&gt;<br /> size((50));<br /> dot((5,0));<br /> dot((5,5));<br /> dot((0,5));<br /> dot((-5,5));<br /> dot((-5,0));<br /> dot((-5,-5));<br /> dot((0,-5));<br /> dot((5,-5));<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7} &lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and &lt;math&gt;\frac{3}{4}&lt;/math&gt; of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12\qquad<br /> \textbf{(B)}\ 17\qquad<br /> \textbf{(C)}\ 24\qquad<br /> \textbf{(D)}\ 27\qquad<br /> \textbf{(E)}\ 36&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Marie cuts a wedge from a &lt;math&gt;6&lt;/math&gt;-cm cylinder of chocolate as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?<br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.65));<br /> real d=90-63.43494882;<br /> draw(ellipse((origin), 2, 4));<br /> fill((0,4)--(0,-4)--(-8,-4)--(-8,4)--cycle, white);<br /> draw(ellipse((-4,0), 2, 4));<br /> draw((0,4)--(-4,4));<br /> draw((0,-4)--(-4,-4));<br /> draw(shift(-2,0)*rotate(-d-5)*ellipse(origin, 1.82, 4.56), linetype(&quot;10 10&quot;));<br /> draw((-4,4)--(-8,4), dashed);<br /> draw((-4,-4)--(-8,-4), dashed);<br /> draw((-4,4.3)--(-4,5));<br /> draw((0,4.3)--(0,5));<br /> draw((-7,4)--(-7,-4), Arrows(5));<br /> draw((-4,4.7)--(0,4.7), Arrows(5));<br /> label(&quot;$8$ cm&quot;, (-7,0), W);<br /> label(&quot;$6$ cm&quot;, (-2,4.7), N);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} 48 \qquad<br /> \textbf{(B)} 75 \qquad<br /> \textbf{(C)}151\qquad<br /> \textbf{(D)}192 \qquad<br /> \textbf{(E)}603&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> For how many positive integer values of &lt;math&gt;n&lt;/math&gt; are both &lt;math&gt;\frac{n}{3}&lt;/math&gt; and &lt;math&gt;3n&lt;/math&gt; three-digit whole numbers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12\qquad<br /> \textbf{(B)}\ 21\qquad<br /> \textbf{(C)}\ 27\qquad<br /> \textbf{(D)}\ 33\qquad<br /> \textbf{(E)}\ 34&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In square &lt;math&gt;ABCE&lt;/math&gt;, &lt;math&gt;AF=2FE&lt;/math&gt; and &lt;math&gt;CD=2DE&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle BFD&lt;/math&gt; to the area of square &lt;math&gt;ABCE&lt;/math&gt;?<br /> &lt;asy&gt;<br /> size((100));<br /> draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);<br /> draw((3,0)--(9,9)--(0,3)--cycle);<br /> dot((3,0));<br /> dot((0,3));<br /> dot((9,9));<br /> dot((0,0));<br /> dot((9,0));<br /> dot((0,9));<br /> label(&quot;$A$&quot;, (0,9), NW);<br /> label(&quot;$B$&quot;, (9,9), NE);<br /> label(&quot;$C$&quot;, (9,0), SE);<br /> label(&quot;$D$&quot;, (3,0), S);<br /> label(&quot;$E$&quot;, (0,0), SW);<br /> label(&quot;$F$&quot;, (0,3), W);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20} &lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> Ten tiles numbered &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;10&lt;/math&gt; are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{10}\qquad<br /> \textbf{(B)}\ \frac{1}{6}\qquad<br /> \textbf{(C)}\ \frac{11}{60}\qquad<br /> \textbf{(D)}\ \frac{1}{5}\qquad<br /> \textbf{(E)}\ \frac{7}{30}&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Mary's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Approximately what percent of the design is black?<br /> <br /> &lt;asy&gt;<br /> real d=320;<br /> pair O=origin;<br /> pair P=O+8*dir(d);<br /> pair A0 = origin;<br /> pair A1 = O+1*dir(d);<br /> pair A2 = O+2*dir(d);<br /> pair A3 = O+3*dir(d);<br /> pair A4 = O+4*dir(d);<br /> pair A5 = O+5*dir(d);<br /> filldraw(Circle(A0, 6), white, black);<br /> filldraw(circle(A1, 5), black, black);<br /> filldraw(circle(A2, 4), white, black);<br /> filldraw(circle(A3, 3), black, black);<br /> filldraw(circle(A4, 2), white, black);<br /> filldraw(circle(A5, 1), black, black);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad&lt;/math&gt;<br /> <br /> [[2008 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2008|before=[[2007 AMC 8 Problems|2007 AMC 8]]|after=[[2009 AMC 8 Problems|2009 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=User:Math101010&diff=85353 User:Math101010 2017-04-21T03:32:46Z <p>Math101010: /* Goals in Mathematics */</p> <hr /> <div>Math101010 joined AoPS on March 7, 2015, shortly after the new community was released. The March 1, 2015 join date in his profile is an error. Note: The only reason that this actually has content is because Kline reported Math101010 for having a &quot;spammy&quot; user page.<br /> <br /> ==Mathematics==<br /> Math101010 enjoys doing mathematics very much. Once, he insisted on doing math instead of spending time with this family. Except he's not all that great at math. Math101010 likes assuming his own gender and referring to himself in third person.<br /> ===School System===<br /> He is taking Prealgebra.<br /> ===On AoPS===<br /> He started the classes Intro to Geometry in March, Intermediate Algebra in June, and AMC 10 problem series in October 2015. He took all intermediate classes by the end of 2016, and just finished calculus in April 2017.<br /> ===Contests===<br /> In 2015, he scored a rating of 21 in state MATHCOUNTS. He scored a 94.5 in the AMC 10 competition, where he was trolled by the MAA. ([[2015 AMC 10A Problems/Problem 14]])<br /> <br /> In 2016, he scored a 115.5 on the AMC 10A and 100.5 on the AMC 10B. He qualified for the AIME through the A contest, where he scored a 5. He still hopes to qualify for National MATHCOUNTS. He went to ARML, and got a 2, which was &lt;math&gt;\frac16&lt;/math&gt; of his team's total score.<br /> <br /> In 2017, he scored a 120 on the AMC 10A and a 9 on the AIME I. He refuses to talk about the AMC 10B. He completely bombed MATHCOUNTS this year.<br /> <br /> ==Goals in Mathematics==<br /> He hopes to qualify for the USA(J)MO someday. For competitions, he hopes to score at least 132 on the AMC 10 and at least 10 on the AIME. He believes this can be done by taking WOOT, but his parents think it is too expensive and too hard. To date (4/20/2017), Math101010 has never solved a USA(J)MO problem without assistance, but hopes to be able to do so soon.<br /> <br /> ==Non-mathematics==<br /> Math101010 officially has no life.<br /> <br /> ==See also==<br /> [[2015 AMC 10A Problems/Problem 14]]<br /> https://www.youtube.com/watch?v=pbcEOD2rRxs<br /> https://artofproblemsolving.com/community/c116925</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=User:Math101010&diff=85352 User:Math101010 2017-04-21T03:16:13Z <p>Math101010: /* Goals in Mathematics */</p> <hr /> <div>Math101010 joined AoPS on March 7, 2015, shortly after the new community was released. The March 1, 2015 join date in his profile is an error. Note: The only reason that this actually has content is because Kline reported Math101010 for having a &quot;spammy&quot; user page.<br /> <br /> ==Mathematics==<br /> Math101010 enjoys doing mathematics very much. Once, he insisted on doing math instead of spending time with this family. Except he's not all that great at math. Math101010 likes assuming his own gender and referring to himself in third person.<br /> ===School System===<br /> He is taking Prealgebra.<br /> ===On AoPS===<br /> He started the classes Intro to Geometry in March, Intermediate Algebra in June, and AMC 10 problem series in October 2015. He took all intermediate classes by the end of 2016, and just finished calculus in April 2017.<br /> ===Contests===<br /> In 2015, he scored a rating of 21 in state MATHCOUNTS. He scored a 94.5 in the AMC 10 competition, where he was trolled by the MAA. ([[2015 AMC 10A Problems/Problem 14]])<br /> <br /> In 2016, he scored a 115.5 on the AMC 10A and 100.5 on the AMC 10B. He qualified for the AIME through the A contest, where he scored a 5. He still hopes to qualify for National MATHCOUNTS. He went to ARML, and got a 2, which was &lt;math&gt;\frac16&lt;/math&gt; of his team's total score.<br /> <br /> In 2017, he scored a 120 on the AMC 10A and a 9 on the AIME I. He refuses to talk about the AMC 10B. He completely bombed MATHCOUNTS this year.<br /> <br /> ==Goals in Mathematics==<br /> He hopes to qualify for the USA(J)MO someday. For competitions, he hopes to score at least 132 on the AMC 10 and at least 10 on the AIME. He believes this can be done by taking WOOT, but his parents think it is too expensive and too hard. To date (4/20/2017), Math101010 has never solved a 21st century USA(J)MO problem, and hopes to be able to do so soon.<br /> <br /> ==Non-mathematics==<br /> Math101010 officially has no life.<br /> <br /> ==See also==<br /> [[2015 AMC 10A Problems/Problem 14]]<br /> https://www.youtube.com/watch?v=pbcEOD2rRxs<br /> https://artofproblemsolving.com/community/c116925</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=User:Math101010&diff=85351 User:Math101010 2017-04-21T03:14:06Z <p>Math101010: </p> <hr /> <div>Math101010 joined AoPS on March 7, 2015, shortly after the new community was released. The March 1, 2015 join date in his profile is an error. Note: The only reason that this actually has content is because Kline reported Math101010 for having a &quot;spammy&quot; user page.<br /> <br /> ==Mathematics==<br /> Math101010 enjoys doing mathematics very much. Once, he insisted on doing math instead of spending time with this family. Except he's not all that great at math. Math101010 likes assuming his own gender and referring to himself in third person.<br /> ===School System===<br /> He is taking Prealgebra.<br /> ===On AoPS===<br /> He started the classes Intro to Geometry in March, Intermediate Algebra in June, and AMC 10 problem series in October 2015. He took all intermediate classes by the end of 2016, and just finished calculus in April 2017.<br /> ===Contests===<br /> In 2015, he scored a rating of 21 in state MATHCOUNTS. He scored a 94.5 in the AMC 10 competition, where he was trolled by the MAA. ([[2015 AMC 10A Problems/Problem 14]])<br /> <br /> In 2016, he scored a 115.5 on the AMC 10A and 100.5 on the AMC 10B. He qualified for the AIME through the A contest, where he scored a 5. He still hopes to qualify for National MATHCOUNTS. He went to ARML, and got a 2, which was &lt;math&gt;\frac16&lt;/math&gt; of his team's total score.<br /> <br /> In 2017, he scored a 120 on the AMC 10A and a 9 on the AIME I. He refuses to talk about the AMC 10B. He completely bombed MATHCOUNTS this year.<br /> <br /> ==Goals in Mathematics==<br /> He hopes to qualify for the USA(J)MO someday. For competitions, he hopes to score at least 132 on the AMC 10 and at least 10 on the AIME. He believes this can be done by taking WOOT, but his parents think it is too expensive and too hard. To date (4/20/2017), Math101010 has never solved either a 21st century USA(J)MO problem, and hopes to be able to do so soon.<br /> <br /> ==Non-mathematics==<br /> Math101010 officially has no life.<br /> <br /> ==See also==<br /> [[2015 AMC 10A Problems/Problem 14]]<br /> https://www.youtube.com/watch?v=pbcEOD2rRxs<br /> https://artofproblemsolving.com/community/c116925</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_3&diff=84540 2017 AIME I Problems/Problem 3 2017-03-09T02:58:05Z <p>Math101010: /* Solution */</p> <hr /> <div>==Problem 3==<br /> For a positive integer &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;d_n&lt;/math&gt; be the units digit of &lt;math&gt;1 + 2 + \dots + n&lt;/math&gt;. Find the remainder when<br /> &lt;cmath&gt;\sum_{n=1}^{2017} d_n&lt;/cmath&gt;is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> We see that &lt;math&gt;d(n)&lt;/math&gt; appears in cycles of &lt;math&gt;20&lt;/math&gt;, adding a total of &lt;math&gt;70&lt;/math&gt; each cycle.<br /> Since &lt;math&gt;\left\lfloor\frac{2017}{20}\right\rfloor=100&lt;/math&gt;, we know that by &lt;math&gt;2017&lt;/math&gt;, there have been &lt;math&gt;100&lt;/math&gt; cycles, or &lt;math&gt;7000&lt;/math&gt; has been added. This can be discarded, as we're just looking for the last three digits.<br /> Adding up the first &lt;math&gt;17&lt;/math&gt; of the cycle of &lt;math&gt;20&lt;/math&gt;, we get that the answer is &lt;math&gt;\boxed{069}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=81938 Gmaas 2016-12-18T18:43:35Z <p>Math101010: /* Known Facts About gmaas */</p> <hr /> <div>=== Known Facts About gmaas === <br /> <br /> - gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard.<br /> <br /> - themoocow is gmass. gmass is gmaas's cousin.<br /> <br /> - Gmass is &lt;math&gt;\bold{not}&lt;/math&gt; Colonel Meow.<br /> <br /> - Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br /> <br /> - He lives with sseraj. <br /> <br /> - He is often overfed (with probability &lt;math&gt;\frac{3972}{7891}&lt;/math&gt;), or malnourished (with probability &lt;math&gt;\frac{3919}{7891}&lt;/math&gt;) by sseraj.<br /> <br /> - He has &lt;cmath&gt;\sum_{k=0}^1 (k+1)(k+2)+5&lt;/cmath&gt; supercars, excluding the Purrari.<br /> <br /> - He is an employee of AoPS.<br /> <br /> - He is a gmaas with yellow fur and white hypnotizing eyes.<br /> <br /> - He was born with a tail that is a completely different color from the rest of his fur.<br /> <br /> - His stare is very hypnotizing and effective at getting table scraps.<br /> <br /> - He sometimes appears several minutes before certain classes start as an admin. <br /> <br /> - He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br /> <br /> - It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br /> <br /> - Actually, he is a cat. He said so. And science also says so. <br /> <br /> - He is very famous now, and mods always talk about him before class starts.<br /> <br /> - His favorite food is AoPS textbooks, because they help him digest problems.<br /> <br /> - Gmaas tends to reside in sseraj's fridge.<br /> <br /> - Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br /> <br /> - The fur of Gmaas can protect him from the harsh conditions of a freezer.<br /> <br /> - Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br /> <br /> - Gmaas is a sage omniscient cat.<br /> <br /> - He is looking for suitable places other than sseraj's fridge to live in.<br /> <br /> - Places where gmaas sightings have happened: <br /> ~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br /> ~Alligator Swamp A 1072 <br /> ~Alligator Swamp B 1073<br /> ~Introduction to Algebra A (1170)<br /> ~Welcome to Panda Town Gate 1076<br /> ~Welcome to Gmaas Town Gate 1221<br /> ~Welcome to Gmaas Town Gate 1125<br /> ~33°01'17.4&quot;N 117°05'40.1&quot;W (Rancho Bernardo Road, San Diego, CA)<br /> ~The other side of the ice in Antarctica<br /> ~Feisty Alligator Swamp 1115<br /> ~Introduction to Geometry 1221 (Taught by sseraj)<br /> ~Introduction to Counting and Probability 1142 <br /> ~Feisty-ish Alligator Swamp 1115 (AGAIN)<br /> ~Intermediate Counting and Probability 1137<br /> ~Intermediate Counting and Probability 1207<br /> ~Posting student surveys<br /> ~USF Castle Walls 1203<br /> ~Dark Lord's Hut 1210<br /> ~AMC 10 Problem Series 1200<br /> ~Intermediate Number Theory 1138<br /> ~Introduction To Number Theory 1204. Date:7/27/16.<br /> ~Algebra B 1112<br /> ~33°81'199.4&quot;N 167°05'45.1&quot;W (Unknown location, please try again later)<br /> ~Ocelot Rainforest 1111<br /> ~Intermediate Counting and Probability 1137 <br /> ~Intermediate Number Theory 1138 (AGAIN)<br /> ~AMC 10 Problem Series 1200 (AGAIN)<br /> ~Cat Gathering #339 1222<br /> ~Cat Gathering #312 1535<br /> ~Introduction to Geometry (1190)<br /> ~Hogwarts Introduction to Transfiguration (...16168)<br /> ~Cat Meow Meow Purr 1110<br /> ~Bilbo's Hobbit Hole 1234<br /> ~http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx<br /> ~Pre-Algebra 1 with sseraj 1343 (Year: 2015)<br /> ~Intermediate Counting and Probability 1137<br /> ~Intermediate Number Theory 1229 (AGAIN)<br /> ~Calculus 1226<br /> <br /> - These have all been designated as the most glorious sections of Aopsland now (except the USF castle walls), but deforestation threatens the wild areas (i.e. Alligator Swamps A&amp;B).<br /> <br /> - Gmaas has also been sighted in Olympiad Geometry 1148.<br /> <br /> - Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br /> <br /> - Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does.<br /> <br /> - Gmaas are often under the disguise of a penguin or cat. Look out for them.<br /> <br /> EDIT: The above disguises are rare for a Gmaas, except the cat.<br /> <br /> - He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br /> <br /> EDIT: He IS an AoPS site admin.<br /> <br /> - If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br /> <br /> - Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br /> <br /> - Gmaas might have been viewing (with a &lt;math&gt;\frac{99.999}{100}&lt;/math&gt; chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday.<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Vu.<br /> <br /> - Gmaas is on the list of &quot;Elusive Creatures.&quot; If you have questions, or want the full list, contact moab33.<br /> <br /> - Gmaas can be summoned using the &lt;math&gt;\tan(90)&lt;/math&gt; ritual. Draw a pentagram and write the numerical value of &lt;math&gt;\tan(90)&lt;/math&gt; in the middle, and he will be summoned.<br /> <br /> - Gmaas's left eye contains the singularity. (Only when everyone in the world blinks at the same time.)<br /> <br /> - Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br /> <br /> - The original owner of Gmaas is gmaas.<br /> <br /> - Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br /> <br /> - It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mu.<br /> <br /> - Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond<br /> <br /> - It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br /> <br /> - Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br /> <br /> - It has been confirmed that gmaas uses gmaal as his email service<br /> <br /> - Gmaas enjoys wearing gmean shorts<br /> <br /> - Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br /> <br /> - Gmaas is well known behind his stage name, Michal Stevens (also known as Vsauce XD), or his page name, Purrshanks.<br /> <br /> - Gmass rekt sseraj at 12:54 june 4, 2016 UTC time zone. And then the Doctor chased him.<br /> <br /> - Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br /> <br /> - Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br /> <br /> - In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br /> <br /> - The TV which was given to the DL was destroyed when the hut was raided by a group of hardcore USFers who did not want modern technology in their age, quote, &quot;We beg gmaas' pardon, but such technology in an age of swords, bows, and magic is rather disturbing.&quot; The raiders were astonished to find the TV whole and running when they next returned in a full-scale attack, and have not touched it since.<br /> <br /> - Gmaas is a Super Duper Uper Cat Time Lord. He has 57843504 regenerations and has used &lt;math&gt;3&lt;/math&gt;. &lt;cmath&gt;9\cdot12\cdot2\cdot267794=57843504&lt;/cmath&gt;. <br /> <br /> - Gmaas loves to eat turnips. At &lt;math&gt;\frac{13}{32}&lt;/math&gt; of the sites he was spotted at, he was seen with a turnip.<br /> <br /> - Gmaas has two tails, one for everyday life, and one for special occasions<br /> <br /> === gmaas in Popular Culture ===<br /> <br /> - Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, &quot;The Adventures of gmaas&quot;.<br /> <br /> - BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,&quot;What are gwali?&quot; the customary answer &quot;This is gwali&quot; is returned. Scientist 5space is now looking into it.<br /> <br /> - Sullymath is also writing a book about Gmaas<br /> <br /> - Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br /> <br /> - Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br /> <br /> - Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Answer_Key&diff=81315 2016 AMC 8 Answer Key 2016-11-23T13:54:50Z <p>Math101010: </p> <hr /> <div>#C<br /> #<br /> #A<br /> #<br /> #<br /> #<br /> #B<br /> #C<br /> #B<br /> #D<br /> #B<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #E<br /> #A<br /> #<br /> #C<br /> #<br /> #A<br /> #B</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Answer_Key&diff=81312 2016 AMC 8 Answer Key 2016-11-23T13:51:31Z <p>Math101010: </p> <hr /> <div>#<br /> #<br /> #A<br /> #<br /> #<br /> #<br /> #B<br /> #C<br /> #<br /> #D<br /> #B<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #C<br /> #<br /> #A<br /> #B</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Answer_Key&diff=81309 2016 AMC 8 Answer Key 2016-11-23T13:50:27Z <p>Math101010: </p> <hr /> <div>#<br /> #<br /> #A<br /> #<br /> #<br /> #<br /> #B<br /> #C<br /> #<br /> #D<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems&diff=81290 2016 AMC 8 Problems 2016-11-23T13:45:08Z <p>Math101010: /* Problem 2 */</p> <hr /> <div>==Problem 1==<br /> <br /> <br /> The longest professional tennis match ever played lasted a total of &lt;math&gt;11&lt;/math&gt; hours and &lt;math&gt;5&lt;/math&gt; minutes. How many minutes was this?<br /> <br /> &lt;math&gt;\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 1|Solution<br /> ]]<br /> <br /> ==Problem 2==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;AD=8&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AD}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AMC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 2|Solution<br /> ]]</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems&diff=81287 2016 AMC 8 Problems 2016-11-23T13:44:52Z <p>Math101010: </p> <hr /> <div>==Problem 1==<br /> <br /> <br /> The longest professional tennis match ever played lasted a total of &lt;math&gt;11&lt;/math&gt; hours and &lt;math&gt;5&lt;/math&gt; minutes. How many minutes was this?<br /> <br /> &lt;math&gt;\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 1|Solution<br /> ]]<br /> <br /> ==Problem 2==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;AD=8&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AD}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AMC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24&lt;/math&gt;<br /> [[2016 AMC 8 Problems/Problem 2|Solution<br /> ]]</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Answer_Key&diff=81277 2016 AMC 8 Answer Key 2016-11-23T13:42:03Z <p>Math101010: Created page with &quot;# # # # # # # # # # # # # # # # # # # # # # # # #&quot;</p> <hr /> <div>#<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #<br /> #</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_9&diff=81273 2016 AMC 8 Problems/Problem 9 2016-11-23T13:38:37Z <p>Math101010: </p> <hr /> <div>What is the sum of the distinct prime integer divisors of &lt;math&gt;2016&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The prime factorization is &lt;math&gt;2016=2^5\times3^2\times7&lt;/math&gt;. Since the problem is only asking us for the distinct prime factors, we have &lt;math&gt;2,3,7&lt;/math&gt;. Their desired sum is then &lt;math&gt;\boxed{\textbf{(B) }12}&lt;/math&gt;.</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_9&diff=81272 2016 AMC 8 Problems/Problem 9 2016-11-23T13:38:15Z <p>Math101010: /* Solution */</p> <hr /> <div>What is the sum of the distinct prime integer divisors of &lt;math&gt;2016&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The prime factorization is &lt;math&gt;2016=2^5\times3^2\times7&lt;/math&gt;. Since the problem is only asking us for the distinct prime factors, we have &lt;math&gt;2,3,7&lt;/math&gt;. Their desired sum is then &lt;math&gt;\boxed{12}&lt;/math&gt;.</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_16&diff=80213 2016 AMC 10B Problems/Problem 16 2016-09-07T03:13:48Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The sum of an infinite geometric series is of the form:<br /> &lt;cmath&gt;\begin{split}<br /> S &amp; = \frac{a_1}{1-r} <br /> \end{split}&lt;/cmath&gt;<br /> where &lt;math&gt;a_1&lt;/math&gt; is the first term and &lt;math&gt;r&lt;/math&gt; is the ratio whose absolute value is less than 1.<br /> <br /> We know that the second term is the first term multiplied by the ratio. <br /> In other words:<br /> &lt;cmath&gt;\begin{split}<br /> a_1 \cdot r &amp; = 1 \\<br /> a_1 &amp; = \frac{1}{r}<br /> \end{split}&lt;/cmath&gt;<br /> <br /> Thus, the sum is the following:<br /> &lt;cmath&gt;\begin{split}<br /> S &amp; = \frac{\frac{1}{r}}{1-r} \\<br /> S &amp; =\frac{1}{r-r^2}<br /> \end{split}&lt;/cmath&gt;<br /> <br /> Since we want the minimum value of this expression, we want the maximum value for the denominator, &lt;math&gt;-r^2&lt;/math&gt; &lt;math&gt;+&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;.<br /> The maximum x-value of a quadratic with negative &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\frac{-b}{2a}&lt;/math&gt;.<br /> &lt;cmath&gt;\begin{split}<br /> r &amp; = \frac{-(1)}{2(-1)} \\<br /> r &amp; = \frac{1}{2} <br /> \end{split}&lt;/cmath&gt;<br /> <br /> Plugging &lt;math&gt;r&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{1}{2}&lt;/math&gt; into the quadratic yields:<br /> &lt;cmath&gt;\begin{split}<br /> S &amp; = \frac{1}{\frac{1}{2} -(\frac{1}{2})^2} \\<br /> S &amp; = \frac{1}{\frac{1}{4}} <br /> \end{split}&lt;/cmath&gt;<br /> <br /> Therefore, the minimum sum of our infinite geometric sequence is &lt;math&gt;\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.<br /> (Solution by akaashp11)<br /> <br /> ==Solution 2==<br /> After observation we realize that in order to minimize our sum &lt;math&gt;\frac{a}{1-r}&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt; being the reciprocal of r. The common ratio &lt;math&gt;r&lt;/math&gt; has to be in the form of &lt;math&gt;1/x&lt;/math&gt; with &lt;math&gt;x&lt;/math&gt; being an integer as anything more than &lt;math&gt;1&lt;/math&gt; divided by &lt;math&gt;x&lt;/math&gt; would give a larger sum than a ratio in the form of &lt;math&gt;1/x&lt;/math&gt;. <br /> <br /> The first term has to be &lt;math&gt;x&lt;/math&gt;, so then in order to minim6tt5r5y,7uerthrtdhdthize the sum, we have minimize &lt;math&gt;x&lt;/math&gt;. <br /> <br /> The smallest possible value for &lt;math&gt;x&lt;/math&gt; such that it is an integer that's greater than &lt;math&gt;1&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;. So our first term is &lt;math&gt;2&lt;/math&gt; and our common ratio is &lt;math&gt;1/2&lt;/math&gt;. Thus the sum is &lt;math&gt;\frac{2}{1/2}&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.<br /> Solution 2 by No_One<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems&diff=80059 2012 AIME II Problems 2016-08-22T13:54:43Z <p>Math101010: Undo revision 80054 by Theboombox77 (talk)</p> <hr /> <div>{{AIME Problems|year=2012|n=II}}<br /> <br /> == Problem 1 ==<br /> Find the number of ordered pairs of positive integer solutions &lt;math&gt;(m, n)&lt;/math&gt; to the equation &lt;math&gt;20m + 12n = 2012&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Two geometric sequences &lt;math&gt;a_1, a_2, a_3, \ldots&lt;/math&gt; and &lt;math&gt;b_1, b_2, b_3, \ldots&lt;/math&gt; have the same common ratio, with &lt;math&gt;a_1 = 27&lt;/math&gt;, &lt;math&gt;b_1=99&lt;/math&gt;, and &lt;math&gt;a_{15}=b_{11}&lt;/math&gt;. Find &lt;math&gt;a_9&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male and two female professors in each department. A committee of six professors is to contain three men and three women and must also contain two professors from each of the three departments. Find the number of possible committees that can be formed subject to these requirements.<br /> <br /> [[2012 AIME II Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> Ana, Bob, and Cao bike at constant rates of &lt;math&gt;8.6&lt;/math&gt; meters per second, &lt;math&gt;6.2&lt;/math&gt; meters per second, and &lt;math&gt;5&lt;/math&gt; meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point &lt;math&gt;D&lt;/math&gt; on the south edge of the field. Cao arrives at point &lt;math&gt;D&lt;/math&gt; at the same time that Ana and Bob arrive at &lt;math&gt;D&lt;/math&gt; for the first time. The ratio of the field's length to the field's width to the distance from point &lt;math&gt;D&lt;/math&gt; to the southeast corner of the field can be represented as &lt;math&gt;p : q : r&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; are positive integers with &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; relatively prime. Find &lt;math&gt;p+q+r&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> In the accompanying figure, the outer square &lt;math&gt;S&lt;/math&gt; has side length &lt;math&gt;40&lt;/math&gt;. A second square &lt;math&gt;S'&lt;/math&gt; of side length &lt;math&gt;15&lt;/math&gt; is constructed inside &lt;math&gt;S&lt;/math&gt; with the same center as &lt;math&gt;S&lt;/math&gt; and with sides parallel to those of &lt;math&gt;S&lt;/math&gt;. From each midpoint of a side of &lt;math&gt;S&lt;/math&gt;, segments are drawn to the two closest vertices of &lt;math&gt;S'&lt;/math&gt;. The result is a four-pointed starlike figure inscribed in &lt;math&gt;S&lt;/math&gt;. The star figure is cut out and then folded to form a pyramid with base &lt;math&gt;S'&lt;/math&gt;. Find the volume of this pyramid.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> <br /> pair S1 = (20, 20), S2 = (-20, 20), S3 = (-20, -20), S4 = (20, -20);<br /> pair M1 = (S1+S2)/2, M2 = (S2+S3)/2, M3=(S3+S4)/2, M4=(S4+S1)/2;<br /> pair Sp1 = (7.5, 7.5), Sp2=(-7.5, 7.5), Sp3 = (-7.5, -7.5), Sp4 = (7.5, -7.5);<br /> <br /> draw(S1--S2--S3--S4--cycle);<br /> draw(Sp1--Sp2--Sp3--Sp4--cycle);<br /> draw(Sp1--M1--Sp2--M2--Sp3--M3--Sp4--M4--cycle);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> [[2012 AIME II Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Let &lt;math&gt;z=a+bi&lt;/math&gt; be the complex number with &lt;math&gt;\vert z \vert = 5&lt;/math&gt; and &lt;math&gt;b &gt; 0&lt;/math&gt; such that the distance between &lt;math&gt;(1+2i)z^3&lt;/math&gt; and &lt;math&gt;z^5&lt;/math&gt; is maximized, and let &lt;math&gt;z^4 = c+di&lt;/math&gt;. Find &lt;math&gt;c+d&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the increasing sequence of positive integers whose binary representation has exactly &lt;math&gt;8&lt;/math&gt; ones. Let &lt;math&gt;N&lt;/math&gt; be the 1000th number in &lt;math&gt;S&lt;/math&gt;. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> The complex numbers &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; satisfy the system &lt;cmath&gt; z + \frac{20i}w = 5+i&lt;/cmath&gt;<br /> &lt;cmath&gt;w+\frac{12i}z = -4+10i &lt;/cmath&gt; Find the smallest possible value of &lt;math&gt;\vert zw\vert^2&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers such that &lt;math&gt;\frac{\sin x}{\sin y} = 3&lt;/math&gt; and &lt;math&gt;\frac{\cos x}{\cos y} = \frac12&lt;/math&gt;. The value of &lt;math&gt;\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac pq&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Find the number of positive integers &lt;math&gt;n&lt;/math&gt; less than &lt;math&gt;1000&lt;/math&gt; for which there exists a positive real number &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;n=x\lfloor x \rfloor&lt;/math&gt;.<br /> <br /> Note: &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Let &lt;math&gt;f_1(x) = \frac23 - \frac3{3x+1}&lt;/math&gt;, and for &lt;math&gt;n \ge 2&lt;/math&gt;, define &lt;math&gt;f_n(x) = f_1(f_{n-1}(x))&lt;/math&gt;. The value of &lt;math&gt;x&lt;/math&gt; that satisfies &lt;math&gt;f_{1001}(x) = x-3&lt;/math&gt; can be expressed in the form &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. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> For a positive integer &lt;math&gt;p&lt;/math&gt;, define the positive integer &lt;math&gt;n&lt;/math&gt; to be &lt;math&gt;p&lt;/math&gt;''-safe'' if &lt;math&gt;n&lt;/math&gt; differs in absolute value by more than &lt;math&gt;2&lt;/math&gt; from all multiples of &lt;math&gt;p&lt;/math&gt;. For example, the set of &lt;math&gt;10&lt;/math&gt;-safe numbers is &lt;math&gt;\{ 3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, \ldots\}&lt;/math&gt;. Find the number of positive integers less than or equal to &lt;math&gt;10,000&lt;/math&gt; which are simultaneously &lt;math&gt;7&lt;/math&gt;-safe, &lt;math&gt;11&lt;/math&gt;-safe, and &lt;math&gt;13&lt;/math&gt;-safe.<br /> <br /> [[2012 AIME II Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;\sqrt{111}&lt;/math&gt;. There are four distinct triangles &lt;math&gt;AD_1E_1&lt;/math&gt;, &lt;math&gt;AD_1E_2&lt;/math&gt;, &lt;math&gt;AD_2E_3&lt;/math&gt;, and &lt;math&gt;AD_2E_4&lt;/math&gt;, each congruent to &lt;math&gt;\triangle ABC&lt;/math&gt;,<br /> with &lt;math&gt;BD_1 = BD_2 = \sqrt{11}&lt;/math&gt;. Find &lt;math&gt;\sum_{k=1}^4(CE_k)^2&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> In a group of nine people each person shakes hands with exactly two of the other people from the group. Let &lt;math&gt;N&lt;/math&gt; be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \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. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2012 AIME II Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=79268 Gmaas 2016-07-12T14:39:03Z <p>Math101010: /* Known Facts About gmaas */</p> <hr /> <div>=== Known Facts About gmaas ===<br /> <br /> - Gmaas is 5space's favorite animal. [http://artofproblemsolving.com/wiki/index.php?title=File:Gmaas2.png (Source)]<br /> <br /> - He lives with sseraj. <br /> <br /> - He is often overfed by sseraj.<br /> <br /> - He is an employee of AoPS and teaches but doesn't get paid.<br /> <br /> - He is a gmaas with yellow fur and white hypnotizing eyes.<br /> <br /> - He was born with a tail that is a completely different color from the rest of his fur.<br /> <br /> - His stare is very hypnotizing and effective at getting table scraps.<br /> <br /> - He sometimes appears about several thousand hours before certain classes, such as Introduction to Algebra B, as an admin.<br /> <br /> - He died from too many Rubik's cubes in an Introduction to Algebra A class.<br /> <br /> - It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br /> <br /> - Actually, he is a cat. He said so.<br /> <br /> - He is very famous now, and mods always talk about him before class starts.<br /> <br /> - His favorite food is AoPS textbooks, because they help him digest problems.<br /> <br /> - Gmaas tends to reside in sseraj's fridge.<br /> <br /> - Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br /> <br /> - The fur of Gmaas can protect him from the harsh conditions of a freezer.<br /> <br /> - Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br /> <br /> -Gmaas is a sage omniscient cat.<br /> <br /> -He is looking for suitable places other than sseraj's fridge to live in.<br /> <br /> - Places where gmaas sightings have happened: <br /> ~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br /> ~Alligator Swamp A 1072 <br /> ~Alligator Swamp B 1073<br /> ~Introduction to Algebra A (1170)<br /> ~Welcome to Panda Town Gate 1076<br /> ~Welcome to Gmaas Town Gate 1221<br /> ~Welcome to Gmaas Town Gate 1125<br /> ~33°01'17.4&quot;N 117°05'40.1&quot;W<br /> ~AoPS<br /> ~The other side of the ice in Antarctica<br /> ~Feisty Alligator Swamp 1115<br /> ~Introduction to Geometry 1221 (Taught by sseraj)<br /> ~Introduction to Counting and Probability 1142 <br /> ~Feisty-ish Alligator Swamp 1115 (AGAIN)<br /> ~Intermediate Counting and Probability 1137<br /> ~Intermediate Counting and Probability 1207<br /> ~Posting student surveys<br /> <br /> - These have all been designated as the most glorious sections of Aopsland now, but deforestation threatens the wild areas (i.e. Alligator Swamps A&amp;B).<br /> <br /> - Gmaas has also been sighted in Olympiad Geometry 1148.<br /> <br /> - Gmaas are often under the disguise of a penguin or cat. Look out for them.<br /> <br /> - In relation to above comment, sseraj is often seen posting a gif of a penguin slapping another penguin in his Intro to Algebra A class. gmaas appears to be the slapper, but sources are not confirmed. What new modern studies show is that sxu seemed to able to correctly predict MinionLA's actions in the Intro to Algebra A class, possibly from being possessed by gmaas.<br /> <br /> - He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br /> <br /> - If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br /> <br /> - Is this the real life? Is this just fantasy? No. This is gmaas, the legend.<br /> <br /> === gmaas in Popular Culture ===<br /> <br /> - Currently, [https://docs.google.com/document/d/1mLa2d_9Qgv4C9cZdThyjA6kSf2ULgwvkVjPVqmsoV2w/edit a book] is being written (by JpusheenS) about the adventures of gmaas. It is aptly titled, &quot;The Adventures of gmaas&quot;.<br /> <br /> -BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,&quot;What are gwali?&quot; the customary answer &quot;This is gwali&quot; is returned. Scientist 5space is now looking into it.<br /> <br /> -Sullymath is also writing a book about Gmaas</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_23&diff=78172 2016 AMC 10B Problems/Problem 23 2016-04-22T03:31:02Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> <br /> In regular hexagon &lt;math&gt;ABCDEF&lt;/math&gt;, points &lt;math&gt;W&lt;/math&gt;, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are chosen on sides &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, &lt;math&gt;\overline{EF}&lt;/math&gt;, and &lt;math&gt;\overline{FA}&lt;/math&gt; respectively, so lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;ZW&lt;/math&gt;, &lt;math&gt;YX&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; are parallel and equally spaced. What is the ratio of the area of hexagon &lt;math&gt;WCXYFZ&lt;/math&gt; to the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> We draw a diagram to make our work easier:<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> Assume that &lt;math&gt;AB&lt;/math&gt; is of length &lt;math&gt;1&lt;/math&gt;. Therefore, the area of &lt;math&gt;ABCDEF&lt;/math&gt; is &lt;math&gt;\frac{3\sqrt 3}2&lt;/math&gt;. To find the area of &lt;math&gt;WCXYFZ&lt;/math&gt;, we draw &lt;math&gt;\overline{CF}&lt;/math&gt;, and find the area of the trapezoids &lt;math&gt;WCFZ&lt;/math&gt; and &lt;math&gt;CXYF&lt;/math&gt;. <br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> draw(F--C--B--E--D--A);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> From this, we know that &lt;math&gt;CF=2&lt;/math&gt;. We also know that the combined heights of the trapezoids is &lt;math&gt;\frac{\sqrt 3}3&lt;/math&gt;, since &lt;math&gt;\overline{ZW}&lt;/math&gt; and &lt;math&gt;\overline{YX}&lt;/math&gt; are equally spaced, and the height of each of the trapezoids is &lt;math&gt;\frac{\sqrt 3}6&lt;/math&gt;. From this, we know &lt;math&gt;\overline{ZW}&lt;/math&gt; and &lt;math&gt;\overline{YX}&lt;/math&gt; are each &lt;math&gt;\frac 13&lt;/math&gt; of the way from &lt;math&gt;\overline{CF}&lt;/math&gt; to &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. We know that these are both equal to &lt;math&gt;\frac 53&lt;/math&gt;.<br /> <br /> We find the area of each of the trapezoids, which both happen to be &lt;math&gt;\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}&lt;/math&gt;, and the combined area is &lt;math&gt;\frac{11\sqrt 3}{18}^{*}&lt;/math&gt;.<br /> <br /> We find that &lt;math&gt;\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}&lt;/math&gt; is equal to &lt;math&gt;\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;^*&lt;/math&gt; At this point, you can answer &lt;math&gt;\textbf{(C)}&lt;/math&gt; and move on with your test.<br /> <br /> ==Solution 2 (a lot faster than Solution 1)==<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> draw(F--C--B--E--D--A);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once u have draw these lines, it's just a matter of counting triangles. There are &lt;math&gt;22&lt;/math&gt; small triangles in hexagon &lt;math&gt;ZWCXYF&lt;/math&gt;, and &lt;math&gt;9 \cdot 6 = 54&lt;/math&gt; small triangles in the whole hexagon. <br /> <br /> Thus, the answer is &lt;math&gt;\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=Aops_font&diff=77722 Aops font 2016-03-20T14:46:11Z <p>Math101010: /* Purpose */</p> <hr /> <div>The Aops font is a font that can be used to make certain special symbols in the forum. You enclose some text inside [aops] and [/aops] and the text is rendered in the aops font.<br /> <br /> = Purpose=<br /> The [aops][/aops] tags allow you to render items from the custom font that AoPS uses -- you'll recognize these icons from the rest of the site.<br /> <br /> For example, to type the symbol for the re-sort button &lt;span class=&quot;aops-font&quot;&gt;|&lt;/span&gt;, you would type <br /> :[aops]|[/aops]<br /> <br /> =Use=<br /> == In the Forums==<br /> To use the font in the forums, simply put [aops] before your text and [/aops] after it. The text will be converted as described in the tables below.<br /> == In the Wiki==<br /> Because bbcode does not work in the AoPS wiki, you need to use the following HTML code:<br /> :&lt;code&gt; &lt;nowiki&gt;&lt;span class=&quot;aops-font&quot;&gt;text here!&lt;/span&gt;&lt;/nowiki&gt;&lt;/code&gt;<br /> <br /> =Characters=<br /> ==Commonly Used==<br /> &lt;div style=&quot;font-size:large;&quot;&gt;<br /> {| class=&quot;wikitable&quot;<br /> ! Symbol<br /> ! What to type<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;Y&lt;/span&gt;<br /> | Y<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;_&lt;/span&gt;<br /> | _<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;f&lt;/span&gt;<br /> | f<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;h&lt;/span&gt;<br /> | h<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;j&lt;/span&gt;<br /> | j<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;3&lt;/span&gt;<br /> | 3<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;i&lt;/span&gt;<br /> | i<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;X&lt;/span&gt;<br /> | X<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;9&lt;/span&gt;<br /> | 9<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;k&lt;/span&gt;<br /> | k<br /> |}<br /> ==Full List==<br /> &lt;div style=&quot;font-size:large;&quot;&gt;<br /> {| class=&quot;wikitable&quot;<br /> ! Character<br /> ! Aops font<br /> ! Character<br /> ! Aops font<br /> |-<br /> | a<br /> | &lt;span class=&quot;aops-font&quot;&gt;a&lt;/span&gt;<br /> | A<br /> | &lt;span class=&quot;aops-font&quot;&gt;A&lt;/span&gt;<br /> |-<br /> | b<br /> | &lt;span class=&quot;aops-font&quot;&gt;b&lt;/span&gt;<br /> | B<br /> | &lt;span class=&quot;aops-font&quot;&gt;B&lt;/span&gt;<br /> |-<br /> | c<br /> | &lt;span class=&quot;aops-font&quot;&gt;c&lt;/span&gt;<br /> | C<br /> | &lt;span class=&quot;aops-font&quot;&gt;C&lt;/span&gt;<br /> |-<br /> | d<br /> | &lt;span class=&quot;aops-font&quot;&gt;d&lt;/span&gt;<br /> | D<br /> | &lt;span class=&quot;aops-font&quot;&gt;D&lt;/span&gt;<br /> |-<br /> | e<br /> | &lt;span class=&quot;aops-font&quot;&gt;e&lt;/span&gt;<br /> | E<br /> | &lt;span class=&quot;aops-font&quot;&gt;E&lt;/span&gt;<br /> |-<br /> | f<br /> | &lt;span class=&quot;aops-font&quot;&gt;f&lt;/span&gt;<br /> | F<br /> | &lt;span class=&quot;aops-font&quot;&gt;F&lt;/span&gt;<br /> |-<br /> | g<br /> | &lt;span class=&quot;aops-font&quot;&gt;g&lt;/span&gt;<br /> | G<br /> | &lt;span class=&quot;aops-font&quot;&gt;G&lt;/span&gt;<br /> |-<br /> | h<br /> | &lt;span class=&quot;aops-font&quot;&gt;h&lt;/span&gt;<br /> | H<br /> | &lt;span class=&quot;aops-font&quot;&gt;H&lt;/span&gt;<br /> |-<br /> | i<br /> | &lt;span class=&quot;aops-font&quot;&gt;i&lt;/span&gt;<br /> | I<br /> | &lt;span class=&quot;aops-font&quot;&gt;I&lt;/span&gt;<br /> |-<br /> | j<br /> | &lt;span class=&quot;aops-font&quot;&gt;j&lt;/span&gt;<br /> | J<br /> | &lt;span class=&quot;aops-font&quot;&gt;J&lt;/span&gt;<br /> |-<br /> | k<br /> | &lt;span class=&quot;aops-font&quot;&gt;k&lt;/span&gt;<br /> | K<br /> | &lt;span class=&quot;aops-font&quot;&gt;K&lt;/span&gt;<br /> |-<br /> | l<br /> | &lt;span class=&quot;aops-font&quot;&gt;l&lt;/span&gt;<br /> | L<br /> | &lt;span class=&quot;aops-font&quot;&gt;L&lt;/span&gt;<br /> |-<br /> | m<br /> | &lt;span class=&quot;aops-font&quot;&gt;m&lt;/span&gt;<br /> | M<br /> | &lt;span class=&quot;aops-font&quot;&gt;M&lt;/span&gt;<br /> |-<br /> | n<br /> | &lt;span class=&quot;aops-font&quot;&gt;n&lt;/span&gt;<br /> | N<br /> | &lt;span class=&quot;aops-font&quot;&gt;N&lt;/span&gt;<br /> |-<br /> | o<br /> | &lt;span class=&quot;aops-font&quot;&gt;o&lt;/span&gt;<br /> | O<br /> | &lt;span class=&quot;aops-font&quot;&gt;O&lt;/span&gt;<br /> |-<br /> | p<br /> | &lt;span class=&quot;aops-font&quot;&gt;p&lt;/span&gt;<br /> | P<br /> | &lt;span class=&quot;aops-font&quot;&gt;P&lt;/span&gt;<br /> |-<br /> | q<br /> | &lt;span class=&quot;aops-font&quot;&gt;q&lt;/span&gt;<br /> | Q<br /> | &lt;span class=&quot;aops-font&quot;&gt;Q&lt;/span&gt;<br /> |-<br /> | r<br /> | &lt;span class=&quot;aops-font&quot;&gt;r&lt;/span&gt;<br /> | R<br /> | &lt;span class=&quot;aops-font&quot;&gt;R&lt;/span&gt;<br /> |-<br /> | s<br /> | &lt;span class=&quot;aops-font&quot;&gt;s&lt;/span&gt;<br /> | S<br /> | &lt;span class=&quot;aops-font&quot;&gt;S&lt;/span&gt;<br /> |-<br /> | t<br /> | &lt;span class=&quot;aops-font&quot;&gt;t&lt;/span&gt;<br /> | T<br /> | &lt;span class=&quot;aops-font&quot;&gt;T&lt;/span&gt;<br /> |-<br /> | u<br /> | &lt;span class=&quot;aops-font&quot;&gt;u&lt;/span&gt;<br /> | U<br /> | &lt;span class=&quot;aops-font&quot;&gt;U&lt;/span&gt;<br /> |-<br /> | v<br /> | &lt;span class=&quot;aops-font&quot;&gt;v&lt;/span&gt;<br /> | V<br /> | &lt;span class=&quot;aops-font&quot;&gt;V&lt;/span&gt;<br /> |-<br /> | w<br /> | &lt;span class=&quot;aops-font&quot;&gt;w&lt;/span&gt;<br /> | W<br /> | &lt;span class=&quot;aops-font&quot;&gt;W&lt;/span&gt;<br /> |-<br /> | x<br /> | &lt;span class=&quot;aops-font&quot;&gt;x&lt;/span&gt;<br /> | X<br /> | &lt;span class=&quot;aops-font&quot;&gt;X&lt;/span&gt;<br /> |-<br /> | y<br /> | &lt;span class=&quot;aops-font&quot;&gt;y&lt;/span&gt;<br /> | Y<br /> | &lt;span class=&quot;aops-font&quot;&gt;Y&lt;/span&gt;<br /> |-<br /> | z<br /> | &lt;span class=&quot;aops-font&quot;&gt;z&lt;/span&gt;<br /> | Z<br /> | &lt;span class=&quot;aops-font&quot;&gt;Z&lt;/span&gt;<br /> |-<br /> | 1<br /> | &lt;span class=&quot;aops-font&quot;&gt;1&lt;/span&gt;<br /> | !<br /> | &lt;span class=&quot;aops-font&quot;&gt;!&lt;/span&gt;<br /> |-<br /> | 2<br /> | &lt;span class=&quot;aops-font&quot;&gt;2&lt;/span&gt;<br /> | @<br /> | &lt;span class=&quot;aops-font&quot;&gt;@&lt;/span&gt;<br /> |-<br /> | 3<br /> | &lt;span class=&quot;aops-font&quot;&gt;3&lt;/span&gt;<br /> | #<br /> | &lt;span class=&quot;aops-font&quot;&gt;#&lt;/span&gt;<br /> |-<br /> | 4<br /> | &lt;span class=&quot;aops-font&quot;&gt;4&lt;/span&gt;<br /> | &amp;#36;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;#36;&lt;/span&gt;<br /> |-<br /> | 5<br /> | &lt;span class=&quot;aops-font&quot;&gt;5&lt;/span&gt;<br /> | %<br /> | &lt;span class=&quot;aops-font&quot;&gt;%&lt;/span&gt;<br /> |-<br /> | 6<br /> | &lt;span class=&quot;aops-font&quot;&gt;6&lt;/span&gt;<br /> | ^<br /> | &lt;span class=&quot;aops-font&quot;&gt;^&lt;/span&gt;<br /> |-<br /> | 7<br /> | &lt;span class=&quot;aops-font&quot;&gt;7&lt;/span&gt;<br /> | &amp;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;&lt;/span&gt;<br /> |-<br /> | 8<br /> | &lt;span class=&quot;aops-font&quot;&gt;8&lt;/span&gt;<br /> | *<br /> | &lt;span class=&quot;aops-font&quot;&gt;*&lt;/span&gt;<br /> |-<br /> | 9<br /> | &lt;span class=&quot;aops-font&quot;&gt;9&lt;/span&gt;<br /> | (<br /> | &lt;span class=&quot;aops-font&quot;&gt;(&lt;/span&gt;<br /> |-<br /> | 0<br /> | &lt;span class=&quot;aops-font&quot;&gt;0&lt;/span&gt;<br /> | )<br /> | &lt;span class=&quot;aops-font&quot;&gt;)&lt;/span&gt;<br /> |-<br /> | -<br /> | &lt;span class=&quot;aops-font&quot;&gt;-&lt;/span&gt;<br /> | _<br /> | &lt;span class=&quot;aops-font&quot;&gt;_&lt;/span&gt;<br /> |-<br /> | =<br /> | &lt;span class=&quot;aops-font&quot;&gt;=&lt;/span&gt; <br /> | +<br /> | &lt;span class=&quot;aops-font&quot;&gt;+&lt;/span&gt; <br /> |-<br /> | ,<br /> | &lt;span class=&quot;aops-font&quot;&gt;,&lt;/span&gt;<br /> | .<br /> | &lt;span class=&quot;aops-font&quot;&gt;.&lt;/span&gt;<br /> |-<br /> | /<br /> | &lt;span class=&quot;aops-font&quot;&gt;/&lt;/span&gt;<br /> | ?<br /> | &lt;span class=&quot;aops-font&quot;&gt;?&lt;/span&gt;<br /> |-<br /> | &lt;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;lt;&lt;/span&gt;<br /> | &gt;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&gt;&lt;/span&gt;<br /> |-<br /> |;<br /> | &lt;span class=&quot;aops-font&quot;&gt;;&lt;/span&gt;<br /> |:<br /> | &lt;span class=&quot;aops-font&quot;&gt;:&lt;/span&gt;<br /> |-<br /> |'<br /> | &lt;span class=&quot;aops-font&quot;&gt;'&lt;/span&gt;<br /> |&quot;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&quot;&lt;/span&gt;<br /> |-<br /> | {<br /> | &lt;span class=&quot;aops-font&quot;&gt;{&lt;/span&gt;<br /> | }<br /> | &lt;span class=&quot;aops-font&quot;&gt;}&lt;/span&gt;<br /> |-<br /> | –<br /> | &lt;span class=&quot;aops-font&quot;&gt;–&lt;/span&gt;<br /> | —<br /> | &lt;span class=&quot;aops-font&quot;&gt;—&lt;/span&gt;<br /> &lt;/div&gt; <br /> [[Category:AoPS forums]]</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=Aops_font&diff=77721 Aops font 2016-03-20T14:44:16Z <p>Math101010: /* Full List */</p> <hr /> <div>The Aops font is a font that can be used to make certain special symbols in the forum. You enclose some text inside [aops] and [/aops] and the text is rendered in the aops font.<br /> <br /> = Purpose=<br /> The [aops][/aops] tags allow you to render items from the custom font that AoPS uses -- you'll recognize these icons from the rest of the site.<br /> <br /> For example, to type the symbol for the edit post button &lt;span class=&quot;aops-font&quot;&gt;L&lt;/span&gt;, you would type <br /> :[aops]L[/aops]<br /> <br /> =Use=<br /> == In the Forums==<br /> To use the font in the forums, simply put [aops] before your text and [/aops] after it. The text will be converted as described in the tables below.<br /> == In the Wiki==<br /> Because bbcode does not work in the AoPS wiki, you need to use the following HTML code:<br /> :&lt;code&gt; &lt;nowiki&gt;&lt;span class=&quot;aops-font&quot;&gt;text here!&lt;/span&gt;&lt;/nowiki&gt;&lt;/code&gt;<br /> <br /> =Characters=<br /> ==Commonly Used==<br /> &lt;div style=&quot;font-size:large;&quot;&gt;<br /> {| class=&quot;wikitable&quot;<br /> ! Symbol<br /> ! What to type<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;Y&lt;/span&gt;<br /> | Y<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;_&lt;/span&gt;<br /> | _<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;f&lt;/span&gt;<br /> | f<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;h&lt;/span&gt;<br /> | h<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;j&lt;/span&gt;<br /> | j<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;3&lt;/span&gt;<br /> | 3<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;i&lt;/span&gt;<br /> | i<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;X&lt;/span&gt;<br /> | X<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;9&lt;/span&gt;<br /> | 9<br /> |-<br /> | &lt;span class=&quot;aops-font&quot;&gt;k&lt;/span&gt;<br /> | k<br /> |}<br /> ==Full List==<br /> &lt;div style=&quot;font-size:large;&quot;&gt;<br /> {| class=&quot;wikitable&quot;<br /> ! Character<br /> ! Aops font<br /> ! Character<br /> ! Aops font<br /> |-<br /> | a<br /> | &lt;span class=&quot;aops-font&quot;&gt;a&lt;/span&gt;<br /> | A<br /> | &lt;span class=&quot;aops-font&quot;&gt;A&lt;/span&gt;<br /> |-<br /> | b<br /> | &lt;span class=&quot;aops-font&quot;&gt;b&lt;/span&gt;<br /> | B<br /> | &lt;span class=&quot;aops-font&quot;&gt;B&lt;/span&gt;<br /> |-<br /> | c<br /> | &lt;span class=&quot;aops-font&quot;&gt;c&lt;/span&gt;<br /> | C<br /> | &lt;span class=&quot;aops-font&quot;&gt;C&lt;/span&gt;<br /> |-<br /> | d<br /> | &lt;span class=&quot;aops-font&quot;&gt;d&lt;/span&gt;<br /> | D<br /> | &lt;span class=&quot;aops-font&quot;&gt;D&lt;/span&gt;<br /> |-<br /> | e<br /> | &lt;span class=&quot;aops-font&quot;&gt;e&lt;/span&gt;<br /> | E<br /> | &lt;span class=&quot;aops-font&quot;&gt;E&lt;/span&gt;<br /> |-<br /> | f<br /> | &lt;span class=&quot;aops-font&quot;&gt;f&lt;/span&gt;<br /> | F<br /> | &lt;span class=&quot;aops-font&quot;&gt;F&lt;/span&gt;<br /> |-<br /> | g<br /> | &lt;span class=&quot;aops-font&quot;&gt;g&lt;/span&gt;<br /> | G<br /> | &lt;span class=&quot;aops-font&quot;&gt;G&lt;/span&gt;<br /> |-<br /> | h<br /> | &lt;span class=&quot;aops-font&quot;&gt;h&lt;/span&gt;<br /> | H<br /> | &lt;span class=&quot;aops-font&quot;&gt;H&lt;/span&gt;<br /> |-<br /> | i<br /> | &lt;span class=&quot;aops-font&quot;&gt;i&lt;/span&gt;<br /> | I<br /> | &lt;span class=&quot;aops-font&quot;&gt;I&lt;/span&gt;<br /> |-<br /> | j<br /> | &lt;span class=&quot;aops-font&quot;&gt;j&lt;/span&gt;<br /> | J<br /> | &lt;span class=&quot;aops-font&quot;&gt;J&lt;/span&gt;<br /> |-<br /> | k<br /> | &lt;span class=&quot;aops-font&quot;&gt;k&lt;/span&gt;<br /> | K<br /> | &lt;span class=&quot;aops-font&quot;&gt;K&lt;/span&gt;<br /> |-<br /> | l<br /> | &lt;span class=&quot;aops-font&quot;&gt;l&lt;/span&gt;<br /> | L<br /> | &lt;span class=&quot;aops-font&quot;&gt;L&lt;/span&gt;<br /> |-<br /> | m<br /> | &lt;span class=&quot;aops-font&quot;&gt;m&lt;/span&gt;<br /> | M<br /> | &lt;span class=&quot;aops-font&quot;&gt;M&lt;/span&gt;<br /> |-<br /> | n<br /> | &lt;span class=&quot;aops-font&quot;&gt;n&lt;/span&gt;<br /> | N<br /> | &lt;span class=&quot;aops-font&quot;&gt;N&lt;/span&gt;<br /> |-<br /> | o<br /> | &lt;span class=&quot;aops-font&quot;&gt;o&lt;/span&gt;<br /> | O<br /> | &lt;span class=&quot;aops-font&quot;&gt;O&lt;/span&gt;<br /> |-<br /> | p<br /> | &lt;span class=&quot;aops-font&quot;&gt;p&lt;/span&gt;<br /> | P<br /> | &lt;span class=&quot;aops-font&quot;&gt;P&lt;/span&gt;<br /> |-<br /> | q<br /> | &lt;span class=&quot;aops-font&quot;&gt;q&lt;/span&gt;<br /> | Q<br /> | &lt;span class=&quot;aops-font&quot;&gt;Q&lt;/span&gt;<br /> |-<br /> | r<br /> | &lt;span class=&quot;aops-font&quot;&gt;r&lt;/span&gt;<br /> | R<br /> | &lt;span class=&quot;aops-font&quot;&gt;R&lt;/span&gt;<br /> |-<br /> | s<br /> | &lt;span class=&quot;aops-font&quot;&gt;s&lt;/span&gt;<br /> | S<br /> | &lt;span class=&quot;aops-font&quot;&gt;S&lt;/span&gt;<br /> |-<br /> | t<br /> | &lt;span class=&quot;aops-font&quot;&gt;t&lt;/span&gt;<br /> | T<br /> | &lt;span class=&quot;aops-font&quot;&gt;T&lt;/span&gt;<br /> |-<br /> | u<br /> | &lt;span class=&quot;aops-font&quot;&gt;u&lt;/span&gt;<br /> | U<br /> | &lt;span class=&quot;aops-font&quot;&gt;U&lt;/span&gt;<br /> |-<br /> | v<br /> | &lt;span class=&quot;aops-font&quot;&gt;v&lt;/span&gt;<br /> | V<br /> | &lt;span class=&quot;aops-font&quot;&gt;V&lt;/span&gt;<br /> |-<br /> | w<br /> | &lt;span class=&quot;aops-font&quot;&gt;w&lt;/span&gt;<br /> | W<br /> | &lt;span class=&quot;aops-font&quot;&gt;W&lt;/span&gt;<br /> |-<br /> | x<br /> | &lt;span class=&quot;aops-font&quot;&gt;x&lt;/span&gt;<br /> | X<br /> | &lt;span class=&quot;aops-font&quot;&gt;X&lt;/span&gt;<br /> |-<br /> | y<br /> | &lt;span class=&quot;aops-font&quot;&gt;y&lt;/span&gt;<br /> | Y<br /> | &lt;span class=&quot;aops-font&quot;&gt;Y&lt;/span&gt;<br /> |-<br /> | z<br /> | &lt;span class=&quot;aops-font&quot;&gt;z&lt;/span&gt;<br /> | Z<br /> | &lt;span class=&quot;aops-font&quot;&gt;Z&lt;/span&gt;<br /> |-<br /> | 1<br /> | &lt;span class=&quot;aops-font&quot;&gt;1&lt;/span&gt;<br /> | !<br /> | &lt;span class=&quot;aops-font&quot;&gt;!&lt;/span&gt;<br /> |-<br /> | 2<br /> | &lt;span class=&quot;aops-font&quot;&gt;2&lt;/span&gt;<br /> | @<br /> | &lt;span class=&quot;aops-font&quot;&gt;@&lt;/span&gt;<br /> |-<br /> | 3<br /> | &lt;span class=&quot;aops-font&quot;&gt;3&lt;/span&gt;<br /> | #<br /> | &lt;span class=&quot;aops-font&quot;&gt;#&lt;/span&gt;<br /> |-<br /> | 4<br /> | &lt;span class=&quot;aops-font&quot;&gt;4&lt;/span&gt;<br /> | &amp;#36;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;#36;&lt;/span&gt;<br /> |-<br /> | 5<br /> | &lt;span class=&quot;aops-font&quot;&gt;5&lt;/span&gt;<br /> | %<br /> | &lt;span class=&quot;aops-font&quot;&gt;%&lt;/span&gt;<br /> |-<br /> | 6<br /> | &lt;span class=&quot;aops-font&quot;&gt;6&lt;/span&gt;<br /> | ^<br /> | &lt;span class=&quot;aops-font&quot;&gt;^&lt;/span&gt;<br /> |-<br /> | 7<br /> | &lt;span class=&quot;aops-font&quot;&gt;7&lt;/span&gt;<br /> | &amp;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;&lt;/span&gt;<br /> |-<br /> | 8<br /> | &lt;span class=&quot;aops-font&quot;&gt;8&lt;/span&gt;<br /> | *<br /> | &lt;span class=&quot;aops-font&quot;&gt;*&lt;/span&gt;<br /> |-<br /> | 9<br /> | &lt;span class=&quot;aops-font&quot;&gt;9&lt;/span&gt;<br /> | (<br /> | &lt;span class=&quot;aops-font&quot;&gt;(&lt;/span&gt;<br /> |-<br /> | 0<br /> | &lt;span class=&quot;aops-font&quot;&gt;0&lt;/span&gt;<br /> | )<br /> | &lt;span class=&quot;aops-font&quot;&gt;)&lt;/span&gt;<br /> |-<br /> | -<br /> | &lt;span class=&quot;aops-font&quot;&gt;-&lt;/span&gt;<br /> | _<br /> | &lt;span class=&quot;aops-font&quot;&gt;_&lt;/span&gt;<br /> |-<br /> | =<br /> | &lt;span class=&quot;aops-font&quot;&gt;=&lt;/span&gt; <br /> | +<br /> | &lt;span class=&quot;aops-font&quot;&gt;+&lt;/span&gt; <br /> |-<br /> | ,<br /> | &lt;span class=&quot;aops-font&quot;&gt;,&lt;/span&gt;<br /> | .<br /> | &lt;span class=&quot;aops-font&quot;&gt;.&lt;/span&gt;<br /> |-<br /> | /<br /> | &lt;span class=&quot;aops-font&quot;&gt;/&lt;/span&gt;<br /> | ?<br /> | &lt;span class=&quot;aops-font&quot;&gt;?&lt;/span&gt;<br /> |-<br /> | &lt;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&amp;lt;&lt;/span&gt;<br /> | &gt;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&gt;&lt;/span&gt;<br /> |-<br /> |;<br /> | &lt;span class=&quot;aops-font&quot;&gt;;&lt;/span&gt;<br /> |:<br /> | &lt;span class=&quot;aops-font&quot;&gt;:&lt;/span&gt;<br /> |-<br /> |'<br /> | &lt;span class=&quot;aops-font&quot;&gt;'&lt;/span&gt;<br /> |&quot;<br /> | &lt;span class=&quot;aops-font&quot;&gt;&quot;&lt;/span&gt;<br /> |-<br /> | {<br /> | &lt;span class=&quot;aops-font&quot;&gt;{&lt;/span&gt;<br /> | }<br /> | &lt;span class=&quot;aops-font&quot;&gt;}&lt;/span&gt;<br /> |-<br /> | –<br /> | &lt;span class=&quot;aops-font&quot;&gt;–&lt;/span&gt;<br /> | —<br /> | &lt;span class=&quot;aops-font&quot;&gt;—&lt;/span&gt;<br /> &lt;/div&gt; <br /> [[Category:AoPS forums]]</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=User:Math101010&diff=77566 User:Math101010 2016-03-16T19:41:15Z <p>Math101010: /* Goals in Mathematics */</p> <hr /> <div>Math101010 joined AoPS on March 7, 2015, shortly after the new community was released. The March 1, 2015 join date in his profile is fake.<br /> <br /> ==Mathematics==<br /> Math101010 enjoys doing mathematics very much. Once, he insisted on doing math instead of spending time with this family. Except he's not all that great at math.<br /> ===School System===<br /> He is taking Algebra 2 G/T at his local high school, once a week, after school.<br /> ===On AoPS===<br /> He started the classes Intro to Geometry in March, Intermediate Algebra in June, and AMC 10 problem series in October 2015. He will start Precalculus in April 2016.<br /> ===Contests===<br /> In 2015, he scored a rating of 21 in state MATHCOUNTS. He scored a 94.5 in the AMC 10 competition, where he was trolled by the MAA. ([[2015 AMC 10A Problems/Problem 14]])<br /> <br /> In 2016, he scored a 115.5 on the AMC 10A and 100.5 on the AMC 10B. He qualified for the AIME through the A contest, where he scored a 5. He still hopes to qualify for National MATHCOUNTS.<br /> <br /> ==Goals in Mathematics==<br /> He hopes to complete Calculus before the end of his freshman year at high school. For competitions, he hopes to score at least 126 on the AMC 10 and at least 102 on the AMC 12.<br /> <br /> ==Non-mathematics==<br /> Math101010 also enjoys doing things that are not math-related. He enjoys music, which includes listening to it, as well as playing in the school wind ensemble, on the alto saxophone. Another thing he enjoys doing is being lazy, because that's the only thing he is good at.<br /> ==See also==<br /> [[2015 AMC 10A Problems/Problem 14]]<br /> <br /> http://artofproblemsolving.com/community/c116925</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=User:Math101010&diff=77441 User:Math101010 2016-03-06T20:53:31Z <p>Math101010: /* Contests */</p> <hr /> <div>Math101010 joined AoPS on March 7, 2015, shortly after the new community was released. The March 1, 2015 join date in his profile is fake.<br /> <br /> ==Mathematics==<br /> Math101010 enjoys doing mathematics very much. Once, he insisted on doing math instead of spending time with this family. Except he's not all that great at math.<br /> ===School System===<br /> He is taking Algebra 2 G/T at his local high school, once a week, after school.<br /> ===On AoPS===<br /> He started the classes Intro to Geometry in March, Intermediate Algebra in June, and AMC 10 problem series in October 2015. He will start Precalculus in April 2016.<br /> ===Contests===<br /> In 2015, he scored a rating of 21 in state MATHCOUNTS. He scored a 94.5 in the AMC 10 competition, where he was trolled by the MAA. ([[2015 AMC 10A Problems/Problem 14]])<br /> <br /> In 2016, he scored a 115.5 on the AMC 10A and 100.5 on the AMC 10B. He qualified for the AIME through the A contest, where he scored a 5. He still hopes to qualify for National MATHCOUNTS.<br /> <br /> ==Goals in Mathematics==<br /> He hopes to complete Calculus before the end of his freshman year at high school. For competitions, he hopes to score at least 108 on the AMC 10 and at least 86 on the AMC 12.<br /> ==Non-mathematics==<br /> Math101010 also enjoys doing things that are not math-related. He enjoys music, which includes listening to it, as well as playing in the school wind ensemble, on the alto saxophone. Another thing he enjoys doing is being lazy, because that's the only thing he is good at.<br /> ==See also==<br /> [[2015 AMC 10A Problems/Problem 14]]<br /> <br /> http://artofproblemsolving.com/community/c116925</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=User:Math101010&diff=77196 User:Math101010 2016-03-03T15:29:46Z <p>Math101010: /* School System */</p> <hr /> <div>Math101010 joined AoPS on March 7, 2015, shortly after the new community was released. The March 1, 2015 join date in his profile is fake.<br /> <br /> ==Mathematics==<br /> Math101010 enjoys doing mathematics very much. Once, he insisted on doing math instead of spending time with this family. Except he's not all that great at math.<br /> ===School System===<br /> He is taking Algebra 2 G/T at his local high school, once a week, after school.<br /> ===On AoPS===<br /> He started the classes Intro to Geometry in March, Intermediate Algebra in June, and AMC 10 problem series in October 2015. He will start Precalculus in April 2016.<br /> ===Contests===<br /> In 2015, he scored a rating of 21 in state MATHCOUNTS. He scored a 94.5 in the AMC 10 competition, where he was trolled by the MAA. ([[2015 AMC 10A Problems/Problem 14]])<br /> <br /> In 2016, he scored a 115.5 on the AMC 10A and 100.5 on the AMC 10B. He qualified for the AIME through the A contest. He still hopes to qualify for National MATHCOUNTS.<br /> <br /> ==Goals in Mathematics==<br /> He hopes to complete Calculus before the end of his freshman year at high school. For competitions, he hopes to score at least 108 on the AMC 10 and at least 86 on the AMC 12.<br /> ==Non-mathematics==<br /> Math101010 also enjoys doing things that are not math-related. He enjoys music, which includes listening to it, as well as playing in the school wind ensemble, on the alto saxophone. Another thing he enjoys doing is being lazy, because that's the only thing he is good at.<br /> ==See also==<br /> [[2015 AMC 10A Problems/Problem 14]]<br /> <br /> http://artofproblemsolving.com/community/c116925</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=User:Math101010&diff=77195 User:Math101010 2016-03-03T15:28:56Z <p>Math101010: /* Contests */</p> <hr /> <div>Math101010 joined AoPS on March 7, 2015, shortly after the new community was released. The March 1, 2015 join date in his profile is fake.<br /> <br /> ==Mathematics==<br /> Math101010 enjoys doing mathematics very much. Once, he insisted on doing math instead of spending time with this family. Except he's not all that great at math.<br /> ==School System==<br /> He is taking Algebra 2 G/T at his local high school, once a week, after school.<br /> ===On AoPS===<br /> He started the classes Intro to Geometry in March, Intermediate Algebra in June, and AMC 10 problem series in October 2015. He will start Precalculus in April 2016.<br /> ===Contests===<br /> In 2015, he scored a rating of 21 in state MATHCOUNTS. He scored a 94.5 in the AMC 10 competition, where he was trolled by the MAA. ([[2015 AMC 10A Problems/Problem 14]])<br /> <br /> In 2016, he scored a 115.5 on the AMC 10A and 100.5 on the AMC 10B. He qualified for the AIME through the A contest. He still hopes to qualify for National MATHCOUNTS.<br /> <br /> ==Goals in Mathematics==<br /> He hopes to complete Calculus before the end of his freshman year at high school. For competitions, he hopes to score at least 108 on the AMC 10 and at least 86 on the AMC 12.<br /> ==Non-mathematics==<br /> Math101010 also enjoys doing things that are not math-related. He enjoys music, which includes listening to it, as well as playing in the school wind ensemble, on the alto saxophone. Another thing he enjoys doing is being lazy, because that's the only thing he is good at.<br /> ==See also==<br /> [[2015 AMC 10A Problems/Problem 14]]<br /> <br /> http://artofproblemsolving.com/community/c116925</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_23&diff=77044 2016 AMC 10B Problems/Problem 23 2016-02-25T23:38:58Z <p>Math101010: /* Solution 2 (30 sec) */</p> <hr /> <div>==Problem==<br /> <br /> In regular hexagon &lt;math&gt;ABCDEF&lt;/math&gt;, points &lt;math&gt;W&lt;/math&gt;, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are chosen on sides &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, &lt;math&gt;\overline{EF}&lt;/math&gt;, and &lt;math&gt;\overline{FA}&lt;/math&gt; respectively, so lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;ZW&lt;/math&gt;, &lt;math&gt;YX&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; are parallel and equally spaced. What is the ratio of the area of hexagon &lt;math&gt;WCXYFZ&lt;/math&gt; to the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> We draw a diagram to make our work easier:<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> Assume that &lt;math&gt;AB&lt;/math&gt; is of length &lt;math&gt;1&lt;/math&gt;. Therefore, the area of &lt;math&gt;ABCDEF&lt;/math&gt; is &lt;math&gt;\frac{3\sqrt 3}2&lt;/math&gt;. To find the area of &lt;math&gt;WCXYFZ&lt;/math&gt;, we draw &lt;math&gt;\overline{CF}&lt;/math&gt;, and find the area of the trapezoids &lt;math&gt;WCFZ&lt;/math&gt; and &lt;math&gt;CXYF&lt;/math&gt;. <br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> draw(F--C--B--E--D--A);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> From this, we know that &lt;math&gt;CF=2&lt;/math&gt;. We also know that the combined heights of the trapezoids is &lt;math&gt;\frac{\sqrt 3}3&lt;/math&gt;, since &lt;math&gt;\overline{ZW}&lt;/math&gt; and &lt;math&gt;\overline{YX}&lt;/math&gt; are equally spaced, and the height of each of the trapezoids is &lt;math&gt;\frac{\sqrt 3}6&lt;/math&gt;. From this, we know &lt;math&gt;\overline{ZW}&lt;/math&gt; and &lt;math&gt;\overline{YX}&lt;/math&gt; are each &lt;math&gt;\frac 13&lt;/math&gt; of the way from &lt;math&gt;\overline{CF}&lt;/math&gt; to &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. We know that these are both equal to &lt;math&gt;\frac 53&lt;/math&gt;.<br /> <br /> We find the area of each of the trapezoids, which both happen to be &lt;math&gt;\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}&lt;/math&gt;, and the combined area is &lt;math&gt;\frac{11\sqrt 3}{18}^{*}&lt;/math&gt;.<br /> <br /> We find that &lt;math&gt;\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}&lt;/math&gt; is equal to &lt;math&gt;\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;^*&lt;/math&gt; At this point, you can answer &lt;math&gt;\textbf{(C)}&lt;/math&gt; and move on with your test.<br /> <br /> ==Solution 2==<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> draw(F--C--B--E--D--A);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once u have draw these lines, it's just a matter of counting triangles. There are &lt;math&gt;22&lt;/math&gt; small triangles in hexagon &lt;math&gt;ZWCXYF&lt;/math&gt;, and &lt;math&gt;9 \cdot 6 = 54&lt;/math&gt; small triangles in the whole hexagon. <br /> <br /> Thus, the answer is &lt;math&gt;\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_23&diff=76991 2016 AMC 10B Problems/Problem 23 2016-02-23T02:18:31Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> <br /> In regular hexagon &lt;math&gt;ABCDEF&lt;/math&gt;, points &lt;math&gt;W&lt;/math&gt;, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are chosen on sides &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, &lt;math&gt;\overline{EF}&lt;/math&gt;, and &lt;math&gt;\overline{FA}&lt;/math&gt; respectively, so lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;ZW&lt;/math&gt;, &lt;math&gt;YX&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; are parallel and equally spaced. What is the ratio of the area of hexagon &lt;math&gt;WCXYFZ&lt;/math&gt; to the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> We draw a diagram to make our work easier:<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> Assume that &lt;math&gt;AB&lt;/math&gt; is of length &lt;math&gt;1&lt;/math&gt;. Therefore, the area of &lt;math&gt;ABCDEF&lt;/math&gt; is &lt;math&gt;\frac{3\sqrt 3}2&lt;/math&gt;. To find the area of &lt;math&gt;WCXYFZ&lt;/math&gt;, we draw &lt;math&gt;\overline{CF}&lt;/math&gt;, and find the area of the trapezoids &lt;math&gt;WCFZ&lt;/math&gt; and &lt;math&gt;CXYF&lt;/math&gt;. <br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z;<br /> A=(0,0);<br /> B=(1,0);<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,sqrt(3));<br /> E=(0,sqrt(3));<br /> F=(-1/2,sqrt(3)/2);<br /> W=(4/3,2sqrt(3)/3);<br /> X=(4/3,sqrt(3)/3);<br /> Y=(-1/3,sqrt(3)/3);<br /> Z=(-1/3,2sqrt(3)/3);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(W--Z);<br /> draw(X--Y);<br /> draw(F--C--B--E--D--A);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,NW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$W$&quot;,W,ENE);<br /> label(&quot;$X$&quot;,X,ESE);<br /> label(&quot;$Y$&quot;,Y,WSW);<br /> label(&quot;$Z$&quot;,Z,WNW);<br /> &lt;/asy&gt;<br /> <br /> From this, we know that &lt;math&gt;CF=2&lt;/math&gt;. We also know that the combined heights of the trapezoids is &lt;math&gt;\frac{\sqrt 3}3&lt;/math&gt;, since &lt;math&gt;\overline{ZW}&lt;/math&gt; and &lt;math&gt;\overline{YX}&lt;/math&gt; are equally spaced, and the height of each of the trapezoids is &lt;math&gt;\frac{\sqrt 3}6&lt;/math&gt;. From this, we know &lt;math&gt;\overline{ZW}&lt;/math&gt; and &lt;math&gt;\overline{YX}&lt;/math&gt; are each &lt;math&gt;\frac 13&lt;/math&gt; of the way from &lt;math&gt;\overline{CF}&lt;/math&gt; to &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. We know that these are both equal to &lt;math&gt;\frac 53&lt;/math&gt;.<br /> <br /> We find the area of each of the trapezoids, which both happen to be &lt;math&gt;\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}&lt;/math&gt;, and the combined area is &lt;math&gt;\frac{11\sqrt 3}{18}^{*}&lt;/math&gt;.<br /> <br /> We find that &lt;math&gt;\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}&lt;/math&gt; is equal to &lt;math&gt;\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;^*&lt;/math&gt; At this point, you can answer &lt;math&gt;\textbf{(C)}&lt;/math&gt; and move on with your test.<br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_19&diff=76802 2016 AMC 10B Problems/Problem 19 2016-02-21T19:34:14Z <p>Math101010: /* Solution */</p> <hr /> <div>==Problem==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=5&lt;/math&gt; and &lt;math&gt;BC=4&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; lies on &lt;math&gt;\overline{AB}&lt;/math&gt; so that &lt;math&gt;EB=1&lt;/math&gt;, point &lt;math&gt;G&lt;/math&gt; lies on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;CG=1&lt;/math&gt;. and point &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\overline{CD}&lt;/math&gt; so that &lt;math&gt;DF=2&lt;/math&gt;. Segments &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; intersect &lt;math&gt;\overline{EF}&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt;, respectively. What is the value of &lt;math&gt;\frac{PQ}{EF}&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;pair A1=(2,0),A2=(4,4);<br /> pair B1=(0,4),B2=(5,1);<br /> pair C1=(5,0),C2=(0,4); <br /> draw(A1--A2);<br /> draw(B1--B2);<br /> draw(C1--C2);<br /> draw((0,0)--B1--(5,4)--C1--cycle);<br /> dot((20/7,12/7));<br /> dot((3.07692307692,2.15384615384));<br /> label(&quot;$Q$&quot;,(3.07692307692,2.15384615384),N);<br /> label(&quot;$P$&quot;,(20/7,12/7),W);<br /> label(&quot;$A$&quot;,(0,4), NW);<br /> label(&quot;$B$&quot;,(5,4), NE);<br /> label(&quot;$C$&quot;,(5,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$F$&quot;,(2,0),S); label(&quot;$G$&quot;,(5,1),E);<br /> label(&quot;$E$&quot;,(4,4),N);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad<br /> \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad<br /> \textbf{(C)}~\frac{9}{82} \qquad<br /> \textbf{(D)}~\frac{10}{91}\qquad<br /> \textbf{(E)}~\frac19&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> &lt;math&gt;\textbf{(D)}~\frac{10}{91}&lt;/math&gt;<br /> <br /> solution by ngeorge<br /> <br /> <br /> yay best solution ever<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=76624 AMC historical results 2016-02-21T16:13:16Z <p>Math101010: /* 2016 */</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 /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor:<br /> <br /> ===AMC 12A===<br /> *Average score:<br /> *AIME floor: <br /> <br /> ===AMC 12B===<br /> *Average score:<br /> *AIME floor:<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff: <br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff:<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: <br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: <br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor:88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 67.61<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> <br /> ===AMC 12A===<br /> *Average score: 66.77<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ===AIME II===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<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 /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <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: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: <br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: <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 /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <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: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<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 /> ==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 /> ==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 /> ==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: 120<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 /> ==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 /> ==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: 64.2<br /> *AIME floor: <br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <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: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1998==<br /> ==1997==<br /> ==1996==<br /> ==1995==<br /> ==1994==<br /> ==1993==<br /> ==1992==<br /> ==1991==<br /> ==1990==<br /> ==1989==<br /> ==1988==<br /> ==1987==<br /> ==1986==<br /> ==1985==<br /> ==1984==<br /> ==1983==<br /> ==1982==<br /> ==1981==<br /> ==1980==<br /> ==1979==<br /> ==1978==<br /> ==1977==<br /> ==1976==<br /> ==1975==<br /> ==1974==<br /> ==1973==<br /> ==1972==<br /> ==1971==<br /> ==1970==<br /> ==1969==<br /> ==1968==<br /> ==1967==<br /> ==1966==<br /> ==1965==<br /> ==1964==<br /> ==1963==<br /> ==1962==<br /> ==1961==<br /> ==1960==<br /> ==1959==<br /> ==1958==<br /> ==1957==<br /> ==1956==<br /> ==1955==<br /> ==1954==<br /> ==1953==<br /> ==1952==<br /> ==1951==<br /> ==1950==<br /> <br /> [[Category:Historical results]]</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_15&diff=75144 2016 AMC 10A Problems/Problem 15 2016-02-03T23:30:55Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> Seven cookies of radius &lt;math&gt;1&lt;/math&gt; inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br /> <br /> &lt;asy&gt;<br /> draw(circle((0,0),3));<br /> draw(circle((0,0),1));<br /> draw(circle((1,sqrt(3)),1));<br /> draw(circle((-1,sqrt(3)),1)); <br /> draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br /> draw(circle((2,0),1)); draw(circle((-2,0),1)); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi&lt;/math&gt;<br /> <br /> ==Solution==<br /> The big cookie has radius &lt;math&gt;3&lt;/math&gt;, since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is &lt;math&gt;2 \pi&lt;/math&gt;. The scrap cookie has this area, so its radius must be &lt;math&gt;\boxed{\textbf{(A) }\sqrt 2}&lt;/math&gt;.</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_22&diff=75138 2016 AMC 10A Problems/Problem 22 2016-02-03T23:24:40Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> ==Solution==<br /> Since the prime factorization of &lt;math&gt;110&lt;/math&gt; is &lt;math&gt;2 \cdot 5 \cdot 11&lt;/math&gt;, we have that the number is equal to &lt;math&gt;2 \cdot 5 \cdot 11 \cdot n^3&lt;/math&gt;. This has &lt;math&gt;2 \cdot 2 \cdot 2=8&lt;/math&gt; factors when &lt;math&gt;n=1&lt;/math&gt;. This needs a multiple of 11 factors, which we can achieve by setting &lt;math&gt;n=2^3&lt;/math&gt;, so we have &lt;math&gt;2^10 \cdot 5 \cdot 8&lt;/math&gt; has &lt;math&gt;44&lt;/math&gt; factors. To achieve the desired &lt;math&gt;110&lt;/math&gt; factors, we need the number of factors to also be divisible by &lt;math&gt;5&lt;/math&gt;, so we can set &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;, so &lt;math&gt;2^10 \cdot 5^4 \cdot 11&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; factors. Therefore, &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;. In order to find the number of factors of &lt;math&gt;81n^4&lt;/math&gt;, we raise this to the fourth power and multiply it by &lt;math&gt;81&lt;/math&gt;, and find the factors of that number. We have &lt;math&gt;3^4 \cdot 2^12 \cdot 5^4&lt;/math&gt;, and this has &lt;math&gt;5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}&lt;/math&gt; factors.</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_23&diff=75135 2016 AMC 10A Problems/Problem 23 2016-02-03T23:17:13Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> A binary operation &lt;math&gt;\diamondsuit&lt;/math&gt; has the properties that &lt;math&gt;a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c&lt;/math&gt; and that &lt;math&gt;a\,\diamondsuit \,a=1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. (Here &lt;math&gt;\cdot&lt;/math&gt; represents multiplication). The solution to the equation &lt;math&gt;2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100&lt;/math&gt; can be written as &lt;math&gt;\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p+q?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601&lt;/math&gt;<br /> ==Solution==<br /> <br /> We see that &lt;math&gt;a \diamond a = 1&lt;/math&gt;, and think of division. Testing, we see that the first condition &lt;math&gt;a \diamond (b \diamond c) = (a \diamond b) \cdot c&lt;/math&gt; is satisfied, because &lt;math&gt;\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c&lt;/math&gt;. Therefore, division is the operation &lt;math&gt;\diamond&lt;/math&gt;. Solving the equation, &lt;math&gt;\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100&lt;/math&gt;, so &lt;math&gt;x=\frac{100}{336} = \frac{25}{84}&lt;/math&gt;, so the answer is &lt;math&gt;25 + 84 = \boxed{109}&lt;/math&gt; (A)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_23&diff=75134 2016 AMC 10A Problems/Problem 23 2016-02-03T23:16:55Z <p>Math101010: </p> <hr /> <div>A binary operation &lt;math&gt;\diamondsuit&lt;/math&gt; has the properties that &lt;math&gt;a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c&lt;/math&gt; and that &lt;math&gt;a\,\diamondsuit \,a=1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. (Here &lt;math&gt;\cdot&lt;/math&gt; represents multiplication). The solution to the equation &lt;math&gt;2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100&lt;/math&gt; can be written as &lt;math&gt;\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p+q?&lt;/math&gt;<br /> &lt;math&gt;\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601&lt;/math&gt;<br /> ==Solution==<br /> <br /> We see that &lt;math&gt;a \diamond a = 1&lt;/math&gt;, and think of division. Testing, we see that the first condition &lt;math&gt;a \diamond (b \diamond c) = (a \diamond b) \cdot c&lt;/math&gt; is satisfied, because &lt;math&gt;\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c&lt;/math&gt;. Therefore, division is the operation &lt;math&gt;\diamond&lt;/math&gt;. Solving the equation, &lt;math&gt;\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100&lt;/math&gt;, so &lt;math&gt;x=\frac{100}{336} = \frac{25}{84}&lt;/math&gt;, so the answer is &lt;math&gt;25 + 84 = \boxed{109}&lt;/math&gt; (A)<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_24&diff=75117 2016 AMC 10A Problems/Problem 24 2016-02-03T23:07:49Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}&lt;/math&gt;. Three of the sides of this quadrilateral have length &lt;math&gt;200&lt;/math&gt;. What is the length of the fourth side?<br /> <br /> &lt;math&gt;\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Math101010 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_25&diff=75116 2016 AMC 10A Problems/Problem 25 2016-02-03T23:07:10Z <p>Math101010: </p> <hr /> <div>==Problem==<br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Math101010