https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Zhang2018&feedformat=atom AoPS Wiki - User contributions [en] 2020-12-05T05:51:15Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=111051 2015 AMC 8 Problems/Problem 25 2019-11-10T02:09:27Z <p>Zhang2018: </p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown.<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the &lt;math&gt;4&lt;/math&gt; big triangles that, together with the inscribed square, fill in the large square. If we zoom in on one of the four triangles, we can see that it is composed of a small unit square in the corner of the large square and two triangles, one smaller than the other. We are going to focus specifically on the smaller of the two triangles. This triangle is similar to the big triangle itself by &lt;math&gt;\mathrm{AA}&lt;/math&gt; similarity(because the two sides of a square are parallel. To prove this fact, draw a diagonal of the square and find congruent triangles). Let the shorter leg of the big triangle be &lt;math&gt;x&lt;/math&gt;; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;, because by symmetry, &lt;math&gt;x &lt; \dfrac52&lt;/math&gt;. Note that the other solution we got, namely, &lt;math&gt;x=\dfrac{5+\sqrt 5} 2&lt;/math&gt;, is the length of the segment &lt;math&gt;5-x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;5-x&lt;/math&gt; together sum to &lt;math&gt;5&lt;/math&gt;, the side of the length of the large square, and similarly, the sum of the solutions is &lt;math&gt;5&lt;/math&gt;. This solution is a result of the symmetry of the problem; if we had set the longer leg of the big triangle to be &lt;math&gt;x&lt;/math&gt;, then we would solve the same quadratic to find the same roots, the only difference being that we take the other root.<br /> <br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> Thus the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> ===Solution 2(Contest Solution)=== <br /> <br /> We draw a square as shown:<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); <br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); <br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); <br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); <br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); <br /> path arc = arc((2.5,4),1.5,0,90); <br /> pair P = intersectionpoint(arc,(0,5)--(5,5)); <br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; <br /> draw(P--Pp--Ppp--Pppp--cycle); <br /> filldraw((1,4)--P--(4,4)--cycle,red);<br /> filldraw((4,4)--Pppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Ppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Pp--(1,4)--cycle,red);<br /> &lt;/asy&gt;<br /> <br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. The area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> <br /> Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> <br /> &lt;math&gt;(4)\frac{x}{2} + (4)\frac{y}{2}&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x+y}{2}\right)&lt;/math&gt;<br /> Remember that &lt;math&gt;x+y=3&lt;/math&gt;, so substituting this in we find that the area of all of the triangles is &lt;math&gt;4\left(\frac{3}{2}\right) = 6&lt;/math&gt;. <br /> The area of the &lt;math&gt;4&lt;/math&gt; unit squares is &lt;math&gt;4&lt;/math&gt;, so the area of the square we need is &lt;math&gt;25- (4+6) = 15\implies \boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_17&diff=110985 2002 AMC 12A Problems/Problem 17 2019-11-08T12:24:04Z <p>Zhang2018: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Several sets of prime numbers, such as &lt;math&gt;\{7,83,421,659\}&lt;/math&gt; use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }193<br /> \qquad<br /> \text{(B) }207<br /> \qquad<br /> \text{(C) }225<br /> \qquad<br /> \text{(D) }252<br /> \qquad<br /> \text{(E) }447<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Neither of the digits &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt; can be a units digit of a prime. Therefore the sum of the set is at least &lt;math&gt;40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207&lt;/math&gt;.<br /> <br /> We can indeed create a set of primes with this sum, for example the following sets work: &lt;math&gt;\{ 41, 67, 89, 2, 3, 5 \}&lt;/math&gt; or &lt;math&gt;\{ 43, 61, 89, 2, 5, 7 \}&lt;/math&gt;.<br /> <br /> Thus the answer is &lt;math&gt;207\implies \boxed{\mathrm{(B)}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2002|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_17&diff=110984 2002 AMC 12A Problems/Problem 17 2019-11-08T12:23:36Z <p>Zhang2018: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Several sets of prime numbers, such as &lt;math&gt;\{7,83,421,659\}&lt;/math&gt; use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;<br /> \text{(A) }193<br /> \qquad<br /> \text{(B) }207<br /> \qquad<br /> \text{(C) }225<br /> \qquad<br /> \text{(D) }252<br /> \qquad<br /> \text{(E) }447<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Neither of the digits &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt; can be a units digit of a prime. Therefore the sum of the set is at least &lt;math&gt;40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207&lt;/math&gt;.<br /> <br /> We can indeed create a set of primes with this sum, for example the following sets work: &lt;math&gt;\{ 41, 67, 89, 2, 3, 5 \}&lt;/math&gt; or &lt;math&gt;\{ 43, 61, 89, 2, 5, 7 \}&lt;/math&gt;.<br /> <br /> Thus the answer is &lt;math&gt;207\implies \boxed{(B)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2002|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_12&diff=110964 2005 AMC 12A Problems/Problem 12 2019-11-07T18:58:28Z <p>Zhang2018: /* Solution */</p> <hr /> <div>== Problem ==<br /> A [[line]] passes through &lt;math&gt;A\ (1,1)&lt;/math&gt; and &lt;math&gt;B\ (100,1000)&lt;/math&gt;. How many other points with integer coordinates are on the line and strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> (\mathrm {A}) \ 0 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 3 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 9<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> For convenience’s sake, we can transform &lt;math&gt;A&lt;/math&gt; to the origin and &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;(99,999)&lt;/math&gt; (this does not change the problem). The line &lt;math&gt;AB&lt;/math&gt; has the [[equation]] &lt;math&gt;y = \frac{999}{99}x = \frac{111}{11}x&lt;/math&gt;. The coordinates are integers if &lt;math&gt;11|x&lt;/math&gt;, so the values of &lt;math&gt;x&lt;/math&gt; are &lt;math&gt;11, 22 \ldots 88&lt;/math&gt;, with a total of &lt;math&gt;8\implies \boxed{\mathrm{(D)}}&lt;/math&gt; coordinates.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=11|num-a=13|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_3&diff=110963 2005 AIME I Problems/Problem 3 2019-11-07T18:26:47Z <p>Zhang2018: </p> <hr /> <div>== Problem ==<br /> How many [[positive integer]]s have exactly three [[proper divisor]]s (positive integral [[divisor]]s excluding itself), each of which is less than 50?<br /> <br /> == Solution ==<br /> Suppose &lt;math&gt;n&lt;/math&gt; is such an [[integer]]. Because &lt;math&gt;n&lt;/math&gt; has &lt;math&gt;3&lt;/math&gt; proper divisors, it must have &lt;math&gt;4&lt;/math&gt; divisors,, so &lt;math&gt;n&lt;/math&gt; must be in the form &lt;math&gt;n=p\cdot q&lt;/math&gt; or &lt;math&gt;n=p^3&lt;/math&gt; for distinct [[prime number]]s &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;. <br /> <br /> In the first case, the three proper divisors of &lt;math&gt;n&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;. Thus, we need to pick two prime numbers less than &lt;math&gt;50&lt;/math&gt;. There are fifteen of these (&lt;math&gt;2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43&lt;/math&gt; and &lt;math&gt;47&lt;/math&gt;) so there are &lt;math&gt; {15 \choose 2} =105&lt;/math&gt; numbers of the first type.<br /> <br /> In the second case, the three proper divisors of &lt;math&gt;n&lt;/math&gt; are 1, &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;p^2&lt;/math&gt;. Thus we need to pick a prime number whose square is less than &lt;math&gt;50&lt;/math&gt;. There are four of these (&lt;math&gt;2, 3, 5&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;) and so four numbers of the second type. <br /> <br /> Thus there are &lt;math&gt;105+4=\boxed{109}&lt;/math&gt; integers that meet the given conditions.<br /> <br /> == See also ==<br /> * [[Divisor_function#Demonstration | Counting divisors of positive integers]]<br /> {{AIME box|year=2005|n=I|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_25&diff=110943 1999 AHSME Problems/Problem 25 2019-11-06T20:29:13Z <p>Zhang2018: </p> <hr /> <div>== Problem ==<br /> There are unique integers &lt;math&gt;a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}&lt;/math&gt; such that<br /> <br /> &lt;cmath&gt;\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}&lt;/cmath&gt;<br /> <br /> where &lt;math&gt;0\leq a_{i} &lt; i&lt;/math&gt; for &lt;math&gt;i = 2,3,\ldots,7&lt;/math&gt;. Find &lt;math&gt;a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12&lt;/math&gt;<br /> <br /> == Solution ==<br /> Multiply out the &lt;math&gt;7!&lt;/math&gt; to get <br /> <br /> &lt;cmath&gt;5 \cdot 6! = (3 \cdot 4 \cdots 7)a_2 + (4 \cdots 7)a_3 + (5 \cdot 6 \cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .&lt;/cmath&gt;<br /> <br /> By [[Wilson's Theorem]] (or by straightforward division), &lt;math&gt;a_7 + 7(a_6 + 6a_5 + \cdots) \equiv 5 \cdot 6! \equiv -5 \equiv 2 \pmod{7}&lt;/math&gt;, so &lt;math&gt;a_7 = 2&lt;/math&gt;. Then we move &lt;math&gt;a_7&lt;/math&gt; to the left and divide through by &lt;math&gt;7&lt;/math&gt; to obtain <br /> <br /> &lt;cmath&gt;\frac{5 \cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.&lt;/cmath&gt; <br /> <br /> We then repeat this procedure &lt;math&gt;\pmod{6}&lt;/math&gt;, from which it follows that &lt;math&gt;a_6 \equiv 514 \equiv 4 \pmod{6}&lt;/math&gt;, and so forth. Continuing, we find the unique solution to be &lt;math&gt;(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)&lt;/math&gt; (uniqueness is assured by the [[Division Theorem]]). The answer is &lt;math&gt;9 \Longrightarrow \boxed{\mathrm{(B)}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1999|num-b=24|num-a=26}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_9&diff=109781 2006 AMC 12B Problems/Problem 9 2019-09-12T19:20:03Z <p>Zhang2018: /* Solution 1 */</p> <hr /> <div>== Problem==<br /> How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order? <br /> <br /> &lt;math&gt;<br /> \text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 7, the ones digit cannot be even).<br /> <br /> If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit.<br /> If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit.<br /> If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,<br /> <br /> and so on.<br /> <br /> So, the answer is &lt;math&gt;3(1+2)+2(3+4)+1(5+6)=\boxed{34} \Rightarrow B&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> The last digit is 4, 6, or 8.<br /> <br /> If the last digit is &lt;math&gt;x&lt;/math&gt;, the possibilities for the first two digits correspond to 2-element subsets of &lt;math&gt;\{1,2,\dots,x-1\}&lt;/math&gt;.<br /> <br /> Thus the answer is &lt;math&gt;{3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = 34&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> The answer must be half of a triangular number (evens and decreasing/increasing) so &lt;math&gt;\boxed{34}&lt;/math&gt; or B.<br /> -zoevv<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_8&diff=109780 2006 AMC 12B Problems/Problem 8 2019-09-12T19:16:25Z <p>Zhang2018: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;The lines &lt;math&gt;x = \frac 14y + a&lt;/math&gt; and &lt;math&gt;y = \frac 14x + b&lt;/math&gt; intersect at the point &lt;math&gt;(1,2)&lt;/math&gt;. What is &lt;math&gt;a + b&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;<br /> \text {(A) } 0 \qquad \text {(B) } \frac 34 \qquad \text {(C) } 1 \qquad \text {(D) } 2 \qquad \text {(E) } \frac 94<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> &lt;math&gt;4x-4a=y&lt;/math&gt;<br /> <br /> &lt;math&gt;4x-4a=\frac{1}{4}x+b&lt;/math&gt;<br /> <br /> &lt;math&gt;4\cdot1-4a=\frac{1}{4}\cdot1+b=2&lt;/math&gt;<br /> <br /> &lt;math&gt;a=\frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;b=\frac{7}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;a+b=\frac{9}{4} \Rightarrow \text{(E)}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Add both equations:<br /> <br /> &lt;math&gt;\begin{cases}x=\frac{1}{4}y+a\\ y=\frac{1}{4}x+b \end{cases}&lt;/math&gt;<br /> <br /> Simplify:<br /> <br /> &lt;math&gt;\frac{4}{4}(x+y)=\frac{1}{4}(x+y)+(a+b)&lt;/math&gt;<br /> <br /> Isolate our solution:<br /> <br /> &lt;math&gt;\frac{3}{4}(x+y)=a+b&lt;/math&gt;<br /> <br /> Substitute the point of intersection &lt;math&gt;[x=1, y=2]&lt;/math&gt;<br /> <br /> &lt;math&gt;a+b=\frac{3}{4}\cdot3=\frac{9}{4} \Rightarrow \text{(E)}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Plugging in &lt;math&gt;(1,2)&lt;/math&gt; into the first equation, and solving for &lt;math&gt;a&lt;/math&gt; we get &lt;math&gt;a&lt;/math&gt; as &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> Doing the same for the second equation for the second equation, we get &lt;math&gt;b&lt;/math&gt; as &lt;math&gt;\frac{7}{4}&lt;/math&gt;<br /> <br /> Adding &lt;math&gt;a+b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4} \Rightarrow \text{(E)}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=105883 AoPS Wiki:FAQ 2019-05-20T21:01:20Z <p>Zhang2018: /* AoPS Acronyms */</p> <hr /> <div>{{shortcut|[[A:FAQ]]}}<br /> <br /> This is a community created list of Frequently Asked Questions about Art of Problem Solving. If you have a request to edit or add a question on this page, please make it [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=416&amp;t=414129 here].<br /> <br /> == General==<br /> <br /> <br /> ==== Can I change my username? ====<br /> <br /> : No, unless you contact Art of Problem Solving. However, note that they can typically only change usernames for privacy reasons (e.g., your current username reveals personal information). <br /> <br /> ==== Why can't I change my avatar? ====<br /> <br /> : You must be a user for two weeks before being able to change your avatar. You can get another avatar by downloading a picture you like and changing the dimensions via computer; however, it's really hard to get a good-looking picture under the size limit. Instead, you could visit one of the many avatar shops to get one (try searching &quot;avatar shop&quot; in other forums on the community page).<br /> <br /> ==== Something looks weird (e.g., blurry, missing line, etc.) ====<br /> <br /> : This is likely due to your browser zoom level. Please make sure your zoom level is at 100% for correct rendering of the web page. If it is at 100% and still doesn't look right, try a refresh or hard refresh. If it still shows up, follow the instructions [https://artofproblemsolving.com/community/c10h1076079_read_me_first__how_to_write_a_bug_report here] to submit a bug report about it in Site Support.<br /> <br /> ==== Can I make more than one account?==== <br /> <br /> :Short answer: No.<br /> <br /> :Long answer: Multiple accounts (multis) are banned on AoPS. Having more than one account leads to issues of not remembering on what account you did what. Using multiple accounts to &quot;game&quot; the system, (e.g. increase rating for posts or in online games) or to attempt to spam without consequences will lead to bans on all accounts associated with you. If you have already made additional accounts, please choose one account and stop using the others.<br /> <br /> ====What software does Art of Problem Solving use to run the website?====<br /> <br /> :* Search: Solr<br /> :* Wiki: MediaWiki<br /> :* Asymptote and &lt;math&gt;\text{\LaTeX}&lt;/math&gt; are generated through their respective binary packages<br /> :* Videos: YouTube<br /> <br /> :All other parts of the website are custom built.<br /> <br /> ====Can you make an AoPS App?====<br /> <br /> :No. Contrary to popular opinion, any app built would be less functional than the website. Any features missing on the website in phone view would also be missing in an app as an app doesn't magically make the screen bigger. In addition, the logistics and costs involved are too great.<br /> <br /> == Forums ==<br /> <br /> ====How do I create a forum?====<br /> <br /> :To create an AoPS forum, a user must be on the AoPS community for at least 2 weeks. To create a forum, hover over the community tab, then click &quot;My AoPS.&quot; You should now see your avatar, and a list of your friends. Now, click on the &quot;My Forums&quot; tab. There, you would be able to see which forum you moderate or administrate, as well as the private forums you can access. Click on the &quot;+&quot; button at the top right. This should lead you to a forum creating page.<br /> <br /> ====Why do some posts say they were posted in the future?====<br /> :The AoPS clock is based on the official Naval Observatory time. This is considered the most accurate time. If your computer's system clock is behind the correct time, recent posts may indicate they were posted in the future. Please correct your computer's clock or enable clock synchronization so that your clock is always correct. Mac users may wish to check [http://www.macinstruct.com/node/92 Synchronize your Mac's Clock With A Time Server]. You can also check [http://www.time.gov the US offical time.]<br /> <br /> ====I've lost admin access to a forum or blog I created, how do I get it back?====<br /> :Please send an email to sheriff@aops.com. They will research your request and restore admin access to your forum if appropriate.<br /> <br /> ====What should I do if I find a glitch in the community?====<br /> :First, search the Site Support forum to check if your issue has already been reported. You can search here.<br /> <br /> :If your issue isn't reported, try refreshing your browser page. Most issues go away after a refresh and there is no need to report the issue unless it continues after you refresh your browser. You can refresh on most browsers in Windows with Ctrl + R and on Mac with Cmd + R. To Hard Refresh, click Shift between Ctrl/Cmd and R.<br /> <br /> :Some commonly known glitches that should not be reported are avatars appearing twice in the topic list or private message, friends appearing more than once in the friend list, the edit icon showing next to a message when it shouldn't, and not being able to search for a forum after going back a page in the community.<br /> <br /> ==== How do I format my post, e.g. bold text, add URLs, etc.? ====<br /> <br /> :AoPS is based on a markup language called BBCode. A tutorial of its functions on AoPS and how to use them can be found [[BBCode:Tutorial|here]].<br /> <br /> ==== How do I hide content in the forums? ====<br /> :Wrap the content you want to hide in [hide] tags.<br /> [hide]Content[/hide]<br /> :If you want to customize the label, instead of saying &quot;Click here to reveal hidden text&quot;, you can do something like:<br /> [hide=Label to display]Content[/hide]<br /> <br /> ====I got the message &quot;Too many messages.&quot; when trying to send a private message, why?====<br /> :To prevent PM spam abuse like &quot;hvbowibvibviorybvirusbshbiuovvisuvib&quot;, users with less than five forum posts are limited to four private messages within a forty-eight hour period.<br /> <br /> ==== If I make more posts, it means I'm a better user, right? ====<br /> :Post quality is far more important than post quantity.<br /> <br /> ==== I have made some posts but my post count did not increase. Why? ====<br /> :When you post in some of the forums, such as the Test Forum, Mafia Forum, Fun Factory, and most user created forums, the post does not count towards your overall post count. And still. Quality over quantity.<br /> <br /> ==== I believe a post needs corrective action. What should I do? ====<br /> :If you believe a post needs moderative action, you may report it by clicking the [aops]z[/aops] icon on the upper-right corner of that post. If it's a minor mistake, you may want to PM the offending user instead and explain how they can make their post better. Usually, you shouldn't publicly post such things on a thread itself, which is called &quot;backseat moderation&quot; and is considered rude and unproductive.<br /> <br /> ==== How do I post images? ====<br /> :While AoPS forums have the ability to attach images, we do not generally recommend doing so, as we can not guarantee the images will be available through upgrades, restorations, etc. We also have limited disk space which causes us to remove attachments from time to time. Therefore, we recommend using a third party image hosting solution. There are many options, but we recommend imgur.com.<br /> <br /> :* Go to imgur.com<br /> :* Click New Post<br /> :* Upload your image<br /> :* Hover over image. A URL, Copy button, and down arrow will appear<br /> :* Click the down arrow<br /> :* Click Get Share Links<br /> :* Click Copy next to the BBCode (Forums) option<br /> :* Paste the coped BBCode into AoPS<br /> <br /> ====How do I become a moderator of a forum?====<br /> :The creators of user-created forums select moderators in several different ways. Contact the creator(s) of the forum for more information.<br /> <br /> :The AoPS staff will select moderators for the major forums on an as-needed basis. If the AoPS staff determines that a forum needs additional moderators, then the staff will reach out to productive users and invite them to become moderators. These are some features that the AoPS staff may consider when selecting new moderators.<br /> <br /> :* An established history of productive posting. No unproductive, spammy, or troll posts in the past several years. (For example, a high school student might have made some poor choices while in elementary school -- but then learned from those mistakes.)<br /> :* No use of \8char or other strategies to get around rules or restrictions on the forums.<br /> :* When the user reports posts, the report is useful and helpful, and the reported post (or thread, or user) really did reflect a serious issue worthy of administrators' attention. The user filled in the &quot;Further details&quot; field with valuable information. The user does not report the use of phrases such as &quot;gosh, golly gee&quot; as profanity.<br /> :* No multis.<br /> :* No backseat moderating, no posting of &quot;\requestlock&quot;.<br /> :* When the user revives an old thread, the new post is valuable and relevant.<br /> :* The user does not use the forums to cheat on Alcumus, homework, contests, or similar activities.<br /> <br /> == Blogs ==<br /> ==== How come I can't create a blog? ====<br /> :One needs to have at least 5 posts in order to make a blog.<br /> <br /> ==== How do I make my blog look nice? ====<br /> :Many AoPSers make their blogs look awesome by applying [[CSS|CSS]], which is a high-level stylesheet language. This can be done by typing CSS code into the CSS box in the Blog Control Panel.<br /> <br /> == Alcumus ==<br /> <br /> ==== How is rating computed? ====<br /> :The rating is more of a prediction of what percentage of problems in the topic the Alcumus engine believes you will get correct. As you get more and more correct, the rating will go up slower and slower. However, if you are predicted to get most correct, and you miss one or two problems, the rating, or prediction of percentage correct, will go down.<br /> <br /> ==== I am stuck on a problem, and changing the topic does not change the problem. ====<br /> :Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before moving on.<br /> <br /> ==== Why can't I change topics? ====<br /> :Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before the topic changes to the currently selected topic.<br /> <br /> == Contests ==<br /> ==== Where can I find past contest questions and solutions? ====<br /> :In the [http://www.artofproblemsolving.com/Forum/resources.php Contests] section.<br /> <br /> ==== How do I get problems onto the contest page? ====<br /> <br /> :Make a topic for each question in the appropriate forum, copy/paste the urls to the National Olympiad. Your problems may eventually be submitted into the Contest page.<br /> <br /> ==== What are the guidelines for posting problems to be added to the contests section? ====<br /> :Refer to the [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=144&amp;t=195579 guidelines in this post].<br /> <br /> ==== Why is the wiki missing many contest questions? ====<br /> :Generally, it is because users have not yet posted them onto the wiki (translation difficulties, not having access to the actual problems, lack of interest, etc). If you have a copy, please post the problems in the Community Section! In some cases, however, problems may be missing due to copyright claims from maths organizations.<br /> <br /> ====Where are all of the mock contests?====<br /> :User-generated mock contests are now in its separate subforum, [https://artofproblemsolving.com/community/c594864_aops_mock_contests AoPS Mock Contests]. Please use that forum for creating, hosting, participating in, or recruiting writers/test solvers for mock contests.<br /> <br /> ====How can I host a mock contest?====<br /> :See posts on [https://artofproblemsolving.com/community/c594864h1168324_hi__im_interested_in_writing_a_mock_contest writing a mock contest] and [https://artofproblemsolving.com/community/c594864h1349281_tips_for_organizing_a_mock_contest Tips for Organizing a Mock Contest].<br /> <br /> == LaTeX and Asymptote ==<br /> ==== What is LaTeX, and how do I use it? ====<br /> <br /> :&lt;math&gt;\text{\LaTeX}&lt;/math&gt; is a typesetting markup language and document preparation system. It is widely used for typesetting expressions containing mathematical formulae. See [https://artofproblemsolving.com/wiki/index.php/LaTeX:About LaTeX:About].<br /> <br /> ==== How can I download LaTeX to use on the forums? ====<br /> <br /> :There are no downloads necessary; the forums and the wiki render LaTeX commands between dollar signs. <br /> <br /> ==== How can I download LaTeX for personal use? ====<br /> :You can download TeXstudio [http://texstudio.sourceforge.net here] or TeXnicCenter [http://www.texniccenter.org here]<br /> <br /> ==== Where can I find a list of LaTeX commands? ====<br /> :See [[LaTeX:Symbols|here]].<br /> <br /> ==== Where can I test LaTeX commands? ====<br /> <br /> :[[A:SAND|Sandbox]] or [http://www.artofproblemsolving.com/Resources/texer.php TeXeR]. You can also use our [http://artofproblemsolving.com/community/c67_test_forum Test Forum].<br /> <br /> ==== Where can I find examples of Asymptote diagrams and code? ====<br /> <br /> :Search this wiki for the &lt;tt&gt;&lt;nowiki&gt;&lt;asy&gt;&lt;/nowiki&gt;&lt;/tt&gt; tag or the Forums for the &lt;tt&gt;&lt;nowiki&gt;[asy]&lt;/nowiki&gt;&lt;/tt&gt; tag. See also [[Asymptote:_Useful_commands_and_their_Output|these examples]] and [[Proofs without words|this article]] (click on the images to obtain the code).<br /> <br /> ==== How can I draw 3D diagrams? ====<br /> <br /> :See [[Asymptote: 3D graphics]].<br /> <br /> ==== What is the cse5 package? ==== <br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=149650 here]. The package contains a set of shorthand commands that implement the behavior of usual commands, for example &lt;tt&gt;D()&lt;/tt&gt; for &lt;tt&gt;draw()&lt;/tt&gt; and &lt;tt&gt;dot()&lt;/tt&gt;, and so forth.<br /> <br /> ==== What is the olympiad package? ====<br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=165767 here]. The package contains a set of commands useful for drawing diagrams related to [[:Category:Olympiad Geometry Problems|olympiad geometry problems]].<br /> <br /> == AoPSWiki ==<br /> ==== Is there a guide for wiki syntax? ====<br /> <br /> :See [http://en.wikipedia.org/wiki/Help:Wiki_markup wiki markup], [[AoPSWiki:Tutorial]], and [[Help:Contents]].<br /> <br /> ==== What do I do if I see a mistake in the wiki? ====<br /> <br /> :Click &quot;edit&quot; and correct the error.<br /> <br /> == Miscellaneous ==<br /> ==== Is it possible to join the AoPS Staff? ====<br /> <br /> :Yes. Mr. Rusczyk will sometimes hire a small army of college students to work as interns, graders, and teaching assistants. You must be at least in your second semester of your senior year and be legal to work in the U.S. (at least 16).<br /> <br /> ==== What is the minimum age to be an assistant in an Art of Problem Solving class? ====<br /> <br /> :You must have graduated from high school, or at least be in the second term of your senior year.<br /> <br /> ==AoPS Acronyms==<br /> *'''AFK'''- Away from keyboard, inactive<br /> *'''AoPS'''- Art of Problem Solving<br /> *'''AIME'''- American Invitational Mathematics Examination<br /> *'''admin(s)'''- Administrator(s)<br /> *'''AMC'''- American Math Competitions<br /> *'''AMO'''-USA Mathematical Olympiad<br /> *'''ATM'''- At the Moment/Automated teller machine<br /> *'''brb'''- Be right Back<br /> *'''BTW'''- By the way<br /> *'''CEMC''' - Centre for Mathematics and Computing<br /> *'''Combo''' - Combinatorics (Counting and Probability)<br /> *'''C&amp;P or C+P or CP''' - Counting and Probability or Contests and Programs<br /> *'''cbrt''' - Cube Root<br /> *'''DHR''' - Distinguished Honor Roll<br /> *'''EBWOP'''- Editing by way of post<br /> *'''FF'''- Fun Factory<br /> *'''FTFY''' - Fixed that for you<br /> *'''FTW'''- For the Win, a game on AoPS<br /> *'''gg'''- Good Game<br /> *'''gj'''- Good Job<br /> *'''glhf'''-Good Luck Have Fun<br /> *'''gtg''' - Got to go<br /> *'''HR''' - Honor Roll<br /> *'''HSM''' - High School Math Forum<br /> *'''ID(R or E)K'''-I Don't (Really or Even) Know<br /> *'''ID(R or E)C'''-I Don't (Really or Even) Care<br /> *'''iff'''-If and only if<br /> *'''IIRC'''- If I recall correctly<br /> *'''IKR'''- I know, right?<br /> *'''IMO'''- In my opinion (or International Math Olympiad, depending on context)<br /> *'''IMHO'''- In my humble/honest opinion <br /> *'''JK'''- Just Kidding<br /> *'''JMO'''- United States of America Junior Mathematical Olympiad<br /> *'''lol'''- Laugh Out Loud<br /> *'''MC'''- Mathcounts, a popular math contest for Middle School students.<br /> *'''NFL'''- Not for long/No friends left/National Football League<br /> *'''NHL'''-National Hockey League<br /> *'''mod(s)'''- Moderator(s)<br /> *'''MOEMS'''- Math Olympiads for Elementary and Middle Schools<br /> *'''MO(S)P'''- Mathematical Olympiad (Summer) Program<br /> *'''MSM'''- Middle School Math Forum<br /> *'''NIMO'''-National Internet Math Olympiad<br /> *'''NT'''- Number Theory<br /> *'''OBC'''- Online by computer<br /> *'''OMG'''- Oh My Gosh<br /> *'''OMO'''-Online Math Open<br /> *'''OP'''- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering<br /> *'''QED'''- Quod erat demonstrandum, Latin for &quot;Which was to be proven&quot;; some English mathematicians use it as an acronym for Quite Elegantly Done<br /> *'''QS&amp;A'''- Questions, Suggestions, and Announcements Forum<br /> *'''ro(t)fl''' - Rolling on the floor laughing<br /> *'''smh''' - Shaking my head/somehow<br /> *'''sqrt''' - Square root<br /> *'''Sticky'''- A post pinned to the top of a forum - a thing you really should read<br /> *'''ToS'''- Terms of Service - a thing you really should read<br /> *'''USA(J)MO'''- USA (Junior) Mathematical Olympiadi<br /> *'''V/LA'''- Vacation or Long Absence/Limited Access<br /> *'''WDYM'''- What Do You Mean?<br /> *'''WIP'''- Work in Progress<br /> *'''WLOG'''- Without loss of generality<br /> *'''WOOT''' - Worldwide Online Olympiad Training<br /> *'''wrt'''- With respect to<br /> *'''wtg''' - Way to go<br /> *'''tytia'''- Thank you, that is all<br /> *'''ty'''- Thank you<br /> *'''ttyl'''- Talk to you later<br /> *'''xD'''- Bursting Laugh<br /> *'''TL;DR'''- Too long; don't read<br /> *'''yw'''- You're welcome<br /> <br /> == FTW! ==<br /> <br /> Please see the [//artofproblemsolving.com/ftw/faq For the Win! FAQ] for many common questions.<br /> <br /> == School ==<br /> <br /> ==== What if I miss a class? ====<br /> :There are classroom transcripts available under My Classes, available at the top right of the web site. You can view these transcripts in order to review any missed material. You can also ask questions on the class message board. Don't worry, though, classroom participation usually isn't weighted heavily.<br /> <br /> ==== Is there audio or video in class? ====<br /> :There is generally no audio or video in the class. The classes are generally text and image based, in an interactive chat room environment, which allows students to ask questions at any time during the class. In addition to audio and video limiting interactivity and being less pedagogically effective, the technology isn't quite there yet for all students to be able to adequately receive streaming audio and video.<br /> <br /> ==== What if I want to drop out of a class? ====<br /> :For any course with more than 2 classes, students can drop the course any time before the third class begins and receive a full refund. No drops are allowed after the third class has started. To drop the class, go to the My Classes section by clicking the My Classes link at the top-right of the website. Then find the area on the right side of the page that lets you drop the class. A refund will be processed within 10 business days.<br /> <br /> ==== For my homework, there is supposed to be a green bar but it's orange, why? ====<br /> <br /> :For the bar to turn green, the writing problem must be attempted. Once it is attempted the bar will turn green.<br /> <br /> ==== I need more time for my homework, what should I do? ====<br /> <br /> :There is a &quot;Request Extension&quot; button in the homework tab of your class. This will automatically extend the due date to 2 days after the normal deadline. If you want more time you need to ask for it in the little comment box, stating the reason why you want the extension, and how much time you want. This request will be looked at by the teachers and they will decide if you get the extension or not. Note that you can only use this button 3 times. After, you will need administrator approvement.<br /> :Otherwise, you can send an email to extensions@aops.com with your username, class name and ID (the number in the class page URL after https://artofproblemsolving.com/class/) and reason for extension. Someone should get back to you within a couple days.</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=American_Regions_Mathematics_League&diff=105778 American Regions Mathematics League 2019-05-12T17:46:03Z <p>Zhang2018: /* History */</p> <hr /> <div>{{WotWAnnounce|week=June 6-12}}<br /> <br /> The '''American Regions Math League''' (ARML) is a [[mathematical problem solving]] competition primarily for U.S. high school students.<br /> <br /> {{Contest Info|name=ARML|region=USA|type=Free Response|difficulty=3.5-6|breakdown=&lt;u&gt;Individual&lt;/u&gt;: 4 (Problem 6/8), 6 (Problem 10)&lt;br&gt;&lt;u&gt;Team&lt;/u&gt;: 3.5 (Problem 1-5), 5 (Problem 6-10)}}<br /> <br /> == How to participate ==<br /> Find out if your area has a team already. Contact ARML for details. If your area does not already have a team, ask ARML how to start one.<br /> <br /> See the [[How to join an ARML team]] wiki page for more info. Teams have 15 students. <br /> <br /> == History ==<br /> ARML was started in 1976 as a maths competition.<br /> <br /> == Format ==<br /> The ARML has several different events:<br /> *A team round, in which the entire team works on ten problems together<br /> *A power round, where the questions are proof-oriented and are in a theme<br /> *An individual round, where each individual takes a ten question test<br /> *Two relay rounds, where the answer to one problem is necessary to solve the next problem<br /> *A super relay, a relay round where each team submits one answer<br /> <br /> == Resources ==<br /> * [http://www.arml.com/ ARML Homepage]<br /> * [http://www.artofproblemsolving.com/Forum/index.php?f=348 ARML Forum]<br /> <br /> === Books ===<br /> There are four books of past ARML problems, though some of them are out of print and/or hard to find.<br /> *[http://www.amazon.com/exec/obidos/ASIN/0962640166/artofproblems-20 ARML-NYSML 1989-1994] (see [[NYSML]]).<br /> * [http://www.arml.com/arml_development/page/index.php?page_type=public&amp;page=16 The ARML book website]<br /> <br /> == See also ==<br /> * [[ARML historical results]]<br /> * [[ARML teams]]<br /> * [[Math contest books]]<br /> * [[Math books]]<br /> * [[Mock ARML]]<br /> <br /> <br /> [[Category:Mathematics competitions]]<br /> [[Category:ARML]]<br /> [[Category:Intermediate mathematics competitions]]</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=Forms_of_Figurative_Language&diff=105569 Forms of Figurative Language 2019-05-02T11:14:43Z <p>Zhang2018: Created page with &quot;Synopsis: Written below is a brief description of figurative language containing a number of examples on how such writing is used in numerous literature works. What is figur...&quot;</p> <hr /> <div>Synopsis: Written below is a brief description of figurative language containing a number of examples on how such writing is used in numerous literature works.<br /> <br /> <br /> What is figurative language?<br /> <br /> Figurative language is an expression that means something else than it literally says. For instance, I tell you &quot;It is raining cats and dogs,&quot; I technically am not trying to communicate the fact that domestic animals are flying down from the sky; I am showing the weather is very rainy and stormy.<br /> <br /> Written below is a brief description of the most frequented forms of figurative language:<br /> <br /> Simile- a phrase comparing two things using &quot;like&quot; or &quot;as&quot;<br /> <br /> Ex: Sally was so tall, she seemed like a tree to many of the people she towered over.<br /> <br /> Metaphor- the comparison of two different objects without using &quot;like&quot; or &quot;as&quot;<br /> <br /> Ex: She was a sunflower: she loved to bask in her own success, and failed to recognize the fact that soon, nighttime would arrive.<br /> <br /> Alliteration- the identical occurence of a letter or sound at the start of a set of words in a set<br /> <br /> Ex: Sally sells seashells.<br /> <br /> Personification- a human quality given to a non-human item[animal, food, etc.]<br /> <br /> Ex: The tree shrieked in agony when Billy accidently barged straight, head-first into the tree.<br /> <br /> Onomatopoeia- a sound written out as a work<br /> <br /> Ex: &quot;Ouch&quot;, &quot;Woof&quot;, &quot;Bark&quot;, &quot;Oink&quot;, &quot;Honk&quot;, etc.<br /> <br /> Idiom- a saying that often has a different communication than what it literally mean<br /> <br /> Ex-&quot;It is raining cats and dogs.&quot;<br /> Ex-&quot;The homework was a piece of cake.&quot;<br /> <br /> Hyperbole- a greatly exaggerated statement beyond the belief to make a point<br /> <br /> Ex: I am so hungry, I could devour a horse.</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=Proof_of_the_Polynomial_Remainder_Theorem&diff=105547 Proof of the Polynomial Remainder Theorem 2019-04-30T22:15:23Z <p>Zhang2018: </p> <hr /> <div>Synopsis: Written below is a brief description of the polynomial remainder theorem. The theorem has a wide range of applications spanning from Algebra to Number Theory. This depicts how important the polynomial remainder theorem truly is, and why it must be taught in all courses and is a great tool.<br /> <br /> The remainder theorem states when a polynomial denoted as &lt;math&gt;f(x)&lt;/math&gt; is divided by &lt;math&gt;x-a&lt;/math&gt; for some value of &lt;math&gt;x&lt;/math&gt;, whether real or unreal, the remainder of &lt;math&gt;\frac{f(x)}{x-a}=f(a)&lt;/math&gt; Written below is the proof of the polynomial remainder theorem.<br /> <br /> All polynomials can be written in the form &lt;math&gt;f(x)=d(x)\cdot{q(x)}+r(x)&lt;/math&gt;, where &lt;math&gt;d(x)&lt;/math&gt; is the divisor of the function/polynomial &lt;math&gt;f(x)&lt;/math&gt;, &lt;math&gt;q(x)&lt;/math&gt; is the quotient. amd &lt;math&gt;r(x)&lt;/math&gt; is the remainder. Because the &lt;math&gt;deg&lt;/math&gt; &lt;math&gt;r=0&lt;/math&gt; or the &lt;math&gt;deg&lt;/math&gt; &lt;math&gt;r&lt;deg&lt;/math&gt; &lt;math&gt;d&lt;/math&gt; and the fact that degrees must be whole numbers(&lt;math&gt;0&lt;/math&gt; and the positive numbers), the &lt;math&gt;deg&lt;/math&gt; &lt;math&gt;r=0&lt;/math&gt;, and so to speak, &lt;math&gt;r(x)&lt;/math&gt; is a constant, which we will denote as &lt;math&gt;b&lt;/math&gt;.<br /> <br /> <br /> Knowing this, we can write<br /> <br /> &lt;math&gt;f(x)=d(x)\cdot{q(x)}+b&lt;/math&gt;<br /> <br /> &lt;math&gt;f(x)=(x-a)\cdot{q(x)}+b&lt;/math&gt;<br /> <br /> &lt;math&gt;f(a)=(a-a)\cdot{q(a)}+b&lt;/math&gt;<br /> <br /> &lt;math&gt;f(a)=b&lt;/math&gt;<br /> <br /> We have hereby proven when the quantity &lt;math&gt;x-a&lt;/math&gt; is divided into a polynomial &lt;math&gt;f(x)&lt;/math&gt; of any degree, the value of &lt;math&gt;f(a)=b&lt;/math&gt;, where b is the remainder. The remainder must be a constant because &lt;math&gt;deg&lt;/math&gt; &lt;math&gt;r=0&lt;/math&gt;.</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=Proof_of_the_Polynomial_Remainder_Theorem&diff=105546 Proof of the Polynomial Remainder Theorem 2019-04-30T21:59:46Z <p>Zhang2018: Synopsis: Written below is a brief description of the polynomial remainder theorem. The theorem has a wide range of application in subjects spanning from Algebra to</p> <hr /> <div>The remainder theorem states when a polynomial denoted as &lt;math&gt;f(x)&lt;/math&gt; is divided by &lt;math&gt;x-a&lt;/math&gt; for some value of &lt;math&gt;x&lt;/math&gt;, whether real or unreal, the remainder of &lt;math&gt;\frac{f(x)}{x-a}=f(a)&lt;/math&gt; Written below is the proof of the polynomial remainder theorem.<br /> <br /> All polynomials can be written in the form &lt;math&gt;f(x)=d(x)\cdot\q(x)+r(x)&lt;/math&gt;, where &lt;math&gt;d(x)&lt;/math&gt; is the divisor of the function/polynomial &lt;math&gt;f(x)&lt;/math&gt;, &lt;math&gt;q(x)&lt;/math&gt; is the quotient. amd &lt;math&gt;r(x)&lt;/math&gt; is the remainder. Because the &lt;math&gt;deg r=0&lt;/math&gt; or &lt;math&gt;deg r\less\deg d&lt;/math&gt; and the &lt;math&gt;deg d=1&lt;/math&gt;, degrees must be whole numbers, and so &lt;math&gt;deg r=0&lt;/math&gt;. So to speak, &lt;math&gt;r(x)&lt;/math&gt; is a constant. We denote this constant &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Knowing this, we can write<br /> <br /> &lt;math&gt;f(x)=d(x)\cdot\q(x)+b&lt;/math&gt;<br /> <br /> &lt;math&gt;f(x)=(x-a)\cdot\q(x)+b&lt;/math&gt;<br /> <br /> &lt;math&gt;f(a)=(a-a)\cdot\q(a)+b&lt;/math&gt;<br /> <br /> &lt;math&gt;f(a)=b&lt;/math&gt;<br /> <br /> We have hereby proven when the quantity &lt;math&gt;x-a&lt;/math&gt; is divided into a polynomial &lt;math&gt;f(x)&lt;/math&gt; of any degree, the value of &lt;math&gt;f(a)=b&lt;/math&gt;, where b is the remainder. The remainder must be a constant because &lt;math&gt;deg r=0&lt;/math&gt;.</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=Circle_Theorems&diff=105537 Circle Theorems 2019-04-29T21:58:30Z <p>Zhang2018: synopsis</p> <hr /> <div>Below is a brief introduction to the &lt;math&gt;10&lt;/math&gt; most fundamental circle theorems.<br /> <br /> Theorem &lt;math&gt;1&lt;/math&gt;: If a line is drawn from the center of a circle to the midpoint of a chord, the lines are perpendicular. Vice versa, if a line is drawn from the center of a circle and is perpendicular to the chord, the point of the intersection is the midpoint of the chord.<br /> <br /> &lt;asy&gt;draw(circle((0,0),5));<br /> draw((0,0)--(0,5));<br /> draw((-3,4)--(3,4));<br /> draw(rightanglemark((0,0),(0,4),(3,4)));<br /> &lt;/asy&gt;<br /> <br /> Theorem &lt;math&gt;2&lt;/math&gt;: If &lt;math&gt;2&lt;/math&gt; radii are drawn from the center of the circle and the points where the &lt;math&gt;2&lt;/math&gt; radii lines intersect the circle meet at another point, the interior degree measure created by the radii is &lt;math&gt;2&lt;/math&gt; times the measure of the interior degree measure of the angle created by the chords. This theorem is rather visual, so refer to the pictures below. Remember the shapes the lines create must be an arrow head or a double triangle, as shown in pictures &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;, respectively.<br /> <br /> <br /> <br /> <br /> <br /> Arrow Head<br /> &lt;asy&gt;draw(circle((0,0),5));<br /> draw((-3.15,-3.883)--(-0.99, 4.901));<br /> draw((2.59, -4.277)--(-0.99, 4.901));<br /> draw((-3.15, -3.883)--(0,0));<br /> draw((2.59, -4.277)--(0,0));<br /> dot((-3.15,-3.883));<br /> label(&quot;A&quot;,(-3.15,-3.883),S);<br /> dot((2.59,-4.277));<br /> label(&quot;B&quot;,(2.59,-4.277),S);<br /> dot((-0.99,4.901));<br /> label(&quot;C&quot;,(-0.99,4.90 1),N);<br /> dot((0,0));<br /> label(&quot;D&quot;,(0,0),N);<br /> markscalefactor=0.08;<br /> draw(anglemark((-3.15,-3.883),(0,0),(2.59,-4.277)));<br /> label(&quot;${2x}^\circ$&quot;,anglemark((-3.15,-3.883),(0,0),(2.59,-4.277)),S,green);<br /> draw(anglemark((-3.15,-3.883),(-0.99,4.901),(2.59,-4.277)));<br /> label(&quot;${x}^\circ$&quot;,anglemark((-3.15,-3.883),(-0.99,4.901),(2.59,-4.277)),SE,green);<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> <br /> <br /> Double Triangle<br /> <br /> &lt;asy&gt;draw(circle((0,0),5));<br /> draw((-3.15,-3.883)--(4.913,-0.929));<br /> draw((2.59, -4.277)--(4.913,-0.929));<br /> draw((-3.15, -3.883)--(0,0));<br /> draw((2.59, -4.277)--(0,0));<br /> dot((-3.15,-3.883));<br /> label(&quot;A&quot;,(-3.15,-3.883),S);<br /> dot((2.59,-4.277));<br /> label(&quot;B&quot;,(2.59,-4.277),S);<br /> dot((4.913,-0.929));<br /> label(&quot;C&quot;,(4.913,-0.929),N);<br /> dot((0,0));<br /> label(&quot;D&quot;,(0,0),N);<br /> markscalefactor=0.08;<br /> draw(anglemark((-3.15,-3.883),(0,0),(2.59,-4.277)));<br /> label(&quot;${2x}^\circ$&quot;,anglemark((-3.15,-3.883),(0,0),(2.59,-4.277)),S,green);<br /> draw(anglemark((-3.15,-3.883),(4.913,-0.929),(2.59,-4.277)));<br /> label(&quot;${x}^\circ$&quot;,anglemark((-3.15,-3.883),(4.913,-0.929),(2.59,-4.277)),SW,green);<br /> &lt;/asy&gt;<br /> <br /> <br /> Theorem &lt;math&gt;3&lt;/math&gt;: If &lt;math&gt;2&lt;/math&gt; radii form a straight line or a &lt;math&gt;180^\circ&lt;/math&gt; angle, this line is the diameter.<br /> <br /> &lt;asy&gt;draw(circle((0,0),5));<br /> draw((-5,0)--(0,0),green);<br /> draw((0,0)--(5,0),red);<br /> dot((-5,0));<br /> label(&quot;A&quot;,(-5,0),W);<br /> dot((0,0));<br /> label(&quot;B&quot;,(0,0),N);<br /> dot((5,0));<br /> label(&quot;C&quot;,(5,0),E);<br /> &lt;/asy&gt;<br /> As seen in the picture above, radii &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; form at straight line or a &lt;math&gt;180^\circ&lt;/math&gt; angle. Therefore, segment &lt;math&gt;AC&lt;/math&gt; is a diameter of &lt;math&gt;\bigodot{B}&lt;/math&gt;.<br /> <br /> Theorem &lt;math&gt;4&lt;/math&gt;:The opposite angles of a cyclic quadrilateral sum to &lt;math&gt;180^\circ&lt;/math&gt;.<br /> <br /> &lt;asy&gt;draw(circle((0,0),5));<br /> &lt;/asy&gt;</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=Proof_of_Quadratic_Theorem&diff=105531 Proof of Quadratic Theorem 2019-04-28T21:45:43Z <p>Zhang2018: Here is a basic proof of the quadratic theorem using the tool of completing the square</p> <hr /> <div>Written below is the proof of the quadratic theorem:<br /> <br /> Quadratics are of the form<br /> <br /> &lt;math&gt;f(x)=ax^2+bx+c=0&lt;/math&gt;<br /> <br /> The roots or zeros of the quadratic are the values in which we can input as the &lt;math&gt;x&lt;/math&gt; value in the function to output &lt;math&gt;0&lt;/math&gt;. The possible &lt;math&gt;x&lt;/math&gt; values that go into a function is referred to as the domain, and the possible &lt;math&gt;y&lt;/math&gt; values that come out of a function are referred to as the range. Be cautious to not confuse this word with the term range in a set of data, or the absolute difference between the greatest and least terms of a set.<br /> <br /> Here is the proof of how to find the roots of a quadratic:<br /> <br /> &lt;math&gt;ax^2+bx+c=0&lt;/math&gt;<br /> <br /> &lt;math&gt;a(x^2+\frac{b}{a}x)+c=0&lt;/math&gt;<br /> <br /> At this point, we use a method called completing the square. <br /> <br /> &lt;math&gt;a(x^2+\frac{b}{a}x)=-c&lt;/math&gt;<br /> <br /> &lt;math&gt;a(x+\frac{b}{2a})^2=-c+\frac{b}{4a}&lt;/math&gt;<br /> <br /> &lt;math&gt;(x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b}{4a^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}&lt;/cmath&gt;<br /> <br /> The proof of this theorem is much more important than the theorem itself. Completing the square, as used above, is an incredibly useful tool and shows up in many competitions. Mastering the use of the tool is an extremely practical skill, and must be learned.</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=Proof_of_Quadratic_Theorem&diff=105530 Proof of Quadratic Theorem 2019-04-28T21:40:23Z <p>Zhang2018: Created page with &quot;Written below is the proof of the quadratic theorem: Quadratics are of the form &lt;math&gt;f(x)=ax^2+bx+c=0&lt;/math&gt; The roots or zeros of the quadratic are the values in which we...&quot;</p> <hr /> <div>Written below is the proof of the quadratic theorem:<br /> <br /> Quadratics are of the form<br /> <br /> &lt;math&gt;f(x)=ax^2+bx+c=0&lt;/math&gt;<br /> <br /> The roots or zeros of the quadratic are the values in which we can input as the &lt;math&gt;x&lt;/math&gt; value in the function to output &lt;math&gt;0&lt;/math&gt;. The possible &lt;math&gt;x&lt;/math&gt; values that go into a function is referred to as the domain, and the possible &lt;math&gt;y&lt;/math&gt; values that come out of a function are referred to as the range. Be cautious to not confuse this word with the term range in a set of data, or the absolute difference between the greatest and least terms of a set.<br /> <br /> Here is the proof of how to find the roots of a quadratic:<br /> <br /> &lt;math&gt;ax^2+bx+c=0<br /> <br /> a(x^2+\frac{b}{a}x)+c=0&lt;/math&gt;<br /> <br /> At this point, we use a method called completing the square. <br /> <br /> &lt;math&gt;a(x^2+\frac{b}{a}x)=-c<br /> <br /> a(x+\brac{b}{2a})^2=-c+\frac{b}{4a}<br /> <br /> (x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b}{4a^2}<br /> <br /> (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}<br /> <br /> x+\frac{b}{2a}=\frac{\pm(b^2-4ac)}{2a}<br /> <br /> x=\frac{-b\pm(b^2-4ac)}{2a}&lt;/math&gt;</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:Competition_ratings&diff=105051 AoPS Wiki:Competition ratings 2019-03-30T17:37:33Z <p>Zhang2018: /* AMC 8 */</p> <hr /> <div>This page contains an approximate estimation of the difficulty level of various [[List of mathematics competitions|competitions]]. It is designed with the intention of introducing contests of similar difficulty levels (but possibly different styles of problems) that readers may like to try to gain more experience.<br /> <br /> Each entry groups the problems into sets of similar difficulty levels and suggests an approximate difficulty rating, on a scale from 1 to 10 (from easiest to hardest). Note that many of these ratings are not directly comparable, because the actual competitions have many different rules; the ratings are generally synchronized with the amount of available time, etc. Also, due to variances within a contest, ranges shown may overlap. A sample problem is provided with each entry, with a link to a solution. <br /> <br /> As you may have guessed with time many competitions got more challenging because many countries got more access to books targeted at olympiad preparation. But especially web site where one can discuss Olympiads such as our very own AoPS!<br /> <br /> If you have some experience with mathematical competitions, we hope that you can help us make the difficulty rankings more accurate. Currently, the system is on a scale from 1 to 10 where 1 is the easiest level, e.g. [http://www.mathlinks.ro/resources.php?c=182&amp;cid=44 early AMC problems] and 10 is hardest level, e.g. [http://www.mathlinks.ro/resources.php?c=37&amp;cid=47 China IMO Team Selection Test.] When considering problem difficulty '''put more emphasis on problem-solving aspects and less so on technical skill requirements'''.<br /> <br /> = Scale =<br /> All levels are estimated and refer to ''averages''. The following is a rough standard based on the USA tier system AMC 8 – AMC 10 – AMC 12 – AIME – USAMO/USAJMO, representing Middle School – Junior High – High School – Challenging High School – Olympiad levels. Other contests can be interpolated against this. <br /> # Problems strictly for beginner, on the easiest elementary school or middle school levels (MOEMS, easy Mathcounts questions, #1-20 on AMC 8s, #1-5 AMC 10s, and others that involve standard techniques introduced up to the middle school level), most traditional middle/high school word problems<br /> # For motivated beginners, harder questions from the previous categories (#21-25 on AMC 8, Challenging Mathcounts questions, #5-20 on AMC 10, #5-10 on AMC 12, the easiest AIME questions, etc), traditional middle/high school word problems with extremely complex problem solving<br /> # Beginner/novice problems that require more creative thinking (MathCounts National, #21-25 on AMC 10, #11-20ish on AMC 12, #1-5 on AIMEs, etc.)<br /> # Intermediate-leveled problems, the most difficult questions on AMC 12s (#21-25s), more difficult AIME-styled questions #6-10.<br /> # Difficult AIME problems (#10-12), simple proof-based problems (JBMO), etc<br /> # High-leveled AIME-styled questions (#13-15). Introductory-leveled Olympiad-level questions (#1,4s).<br /> # Tougher Olympiad-level questions, #1,4s that require more technical knowledge than new students to Olympiad-type questions have, easier #2,5s, etc.<br /> # High-level difficult Olympiad-level questions, eg #2,5s on difficult Olympiad contest and easier #3,6s, etc.<br /> # Expert Olympiad-level questions, eg #3,6s on difficult Olympiad contests.<br /> # Super Expert problems, problems occasionally even unsuitable for very hard competitions (like the IMO) due to being exceedingly tedious/long/difficult (e.g. very few students are capable of solving, even on a worldwide basis).<br /> <br /> = Competitions =<br /> <br /> ==Introductory Competitions==<br /> Most middle school and first-stage high school competitions would fall under this category. Problems in these competitions are usually ranked from 1 to 3. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AIntroductory+mathematics+competitions here].<br /> <br /> === [[MOEMS]] ===<br /> *Division E: '''1'''<br /> *: ''The whole number &lt;math&gt;N&lt;/math&gt; is divisible by &lt;math&gt;7&lt;/math&gt;. &lt;math&gt;N&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;2,3,4,&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. What is the smallest value that &lt;math&gt;N&lt;/math&gt; can be?'' ([http://www.moems.org/sample_files/SampleE.pdf Solution])<br /> *Division M: '''1'''<br /> *: ''The value of a two-digit number is &lt;math&gt;10&lt;/math&gt; times more than the sum of its digits. The units digit is 1 more than twice the tens digit. Find the two-digit number.'' ([http://www.moems.org/sample_files/SampleM.pdf Solution])<br /> <br /> === [[AMC 8]] ===<br /> <br /> * Problem 1 - Problem 12: '''2''' <br /> *: ''What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?'' ([[1989 AJHSME Problems/Problem 10|Solution]])<br /> * Problem 13 - Problem 25: '''2'''<br /> *: ''A fifth number, &lt;math&gt;n&lt;/math&gt;, is added to the set &lt;math&gt;\{ 3,6,9,10 \}&lt;/math&gt; to make the mean of the set of five numbers equal to its median. What is the number of possible values of &lt;math&gt;n&lt;/math&gt;? '' ([[1988 AJHSME Problems/Problem 21|Solution]])<br /> <br /> === [[Mathcounts]] ===<br /> <br /> * Countdown: '''0.5''' (School, Chapter), '''1''' (State, National)<br /> * Sprint: '''1-1.5''' (school), '''1.5''' (Chapter),'''2''' (State), '''2-2.5''' (National)<br /> * Target: '''1.5''' (school), '''2''' (Chapter), '''2-2.5''' (State), '''2.5''' (National)<br /> <br /> === [[AMC 10]] ===<br /> <br /> * Problem 1 - 5: '''1'''<br /> *: ''A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?'' ([[2016 AMC 10A Problems/Problem 5|Solution]])<br /> * Problem 6 - 20: '''2'''<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?'' ([[2012 AMC 10A Problems/Problem 16|Solution]])<br /> * Problem 21 - 25: '''3'''<br /> *: ''The vertices of an equilateral triangle lie on the hyperbola &lt;math&gt;xy=1&lt;/math&gt;, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?'' ([[2017 AMC 10B Problems/Problem 24|Solution]])<br /> <br /> ===[[CEMC|CEMC Multiple Choice Tests]]===<br /> This covers the CEMC Gauss, Pascal, Cayley, and Fermat tests.<br /> <br /> * Part A: '''0.5-1.5'''<br /> *: ''How many different 3-digit whole numbers can be formed using the digits 4, 7, and 9, assuming that no digit can be repeated in a number?'' (2015 Gauss 7 Problem 10)<br /> * Part B: '''1-2'''<br /> *: ''Two lines with slopes &lt;math&gt;\tfrac14&lt;/math&gt; and &lt;math&gt;\tfrac54&lt;/math&gt; intersect at &lt;math&gt;(1,1)&lt;/math&gt;. What is the area of the triangle formed by these two lines and the vertical line &lt;math&gt;x = 5&lt;/math&gt;?'' (2017 Cayley Problem 19)<br /> * Part C (Gauss/Pascal): '''2-2.5'''<br /> *: ''Suppose that &lt;math&gt;\tfrac{2009}{2014} + \tfrac{2019}{n} = \tfrac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;n&lt;/math&gt; are positive integers with &lt;math&gt;\tfrac{a}{b}&lt;/math&gt; in lowest terms. What is the sum of the digits of the smallest positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;a&lt;/math&gt; is a multiple of 1004?'' (2014 Pascal Problem 25)<br /> * Part C (Cayley/Fermat): '''2.5-3'''<br /> *: ''Wayne has 3 green buckets, 3 red buckets, 3 blue buckets, and 3 yellow buckets. He randomly distributes 4 hockey pucks among the green buckets, with each puck equally likely to be put in each bucket. Similarly, he distributes 3 pucks among the red buckets, 2 pucks among the blue buckets, and 1 puck among the yellow buckets. Once he is ﬁnished, what is the probability that a green bucket contains more pucks than each of the other 11 buckets?'' (2018 Fermat Problem 24)<br /> <br /> ===[[CEMC|CEMC Fryer/Galois/Hypatia]]===<br /> <br /> * Problem 1-2: '''1-2'''<br /> * Problem 3-4 (early parts): '''2-3'''<br /> * Problem 3-4 (later parts): '''3-5'''<br /> <br /> ==Intermediate Competitions==<br /> This category consists of all the non-proof math competitions for the middle stages of high school. The difficulty range would normally be from 3 to 6. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AIntermediate+mathematics+competitions here].<br /> <br /> === [[AMC 12]] ===<br /> <br /> * Problem 1-10: '''1.75'''<br /> *: ''A solid box is &lt;math&gt;15&lt;/math&gt; cm by &lt;math&gt;10&lt;/math&gt; cm by &lt;math&gt;8&lt;/math&gt; cm. A new solid is formed by removing a cube &lt;math&gt;3&lt;/math&gt; cm on a side from each corner of this box. What percent of the original volume is removed?'' ([[2003 AMC 12A Problems/Problem 3|Solution]])<br /> * Problem 11-20: '''2.5'''<br /> *: ''An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?'' ([[2006 AMC 12B Problems/Problem 18|Solution]])<br /> * Problem 21-25: '''3.75'''<br /> *: ''Functions &lt;math&gt;f&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; are quadratic, &lt;math&gt;g(x) = - f(100 - x)&lt;/math&gt;, and the graph of &lt;math&gt;g&lt;/math&gt; contains the vertex of the graph of &lt;math&gt;f&lt;/math&gt;. The four &lt;math&gt;x&lt;/math&gt;-intercepts on the two graphs have &lt;math&gt;x&lt;/math&gt;-coordinates &lt;math&gt;x_1&lt;/math&gt;, &lt;math&gt;x_2&lt;/math&gt;, &lt;math&gt;x_3&lt;/math&gt;, and &lt;math&gt;x_4&lt;/math&gt;, in increasing order, and &lt;math&gt;x_3 - x_2 = 150&lt;/math&gt;. The value of &lt;math&gt;x_4 - x_1&lt;/math&gt; is &lt;math&gt;m + n\sqrt p&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. What is &lt;math&gt;m + n + p&lt;/math&gt;?'' ([[2009 AMC 12A Problems/Problem 23|Solution]])<br /> <br /> === [[AIME]] ===<br /> <br /> * Problem 1 - 5: '''3'''<br /> *: Starting at &lt;math&gt;(0,0),&lt;/math&gt; an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let &lt;math&gt;p&lt;/math&gt; be the probability that the object reaches &lt;math&gt;(2,2)&lt;/math&gt; in six or fewer steps. Given that &lt;math&gt;p&lt;/math&gt; can be written in the form &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n.&lt;/math&gt; ([[1995 AIME Problems/Problem 3|Solution]])<br /> * Problem 6 - 10: '''4''' <br /> *: ''Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB=21&lt;/math&gt;, &lt;math&gt;AC=22&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are located on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; is [[parallel]] to &lt;math&gt;\overline{BC}&lt;/math&gt; and contains the center of the inscribed circle of triangle &lt;math&gt;ABC&lt;/math&gt;. Then &lt;math&gt;DE=m/n&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.'' ([[2001 AIME I Problems/Problem 7|Solution]])<br /> * Problem 10 - 12: '''5'''<br /> *: Let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=2014&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the polygon in the complex plane whose vertices are &lt;math&gt;z&lt;/math&gt; and every &lt;math&gt;w&lt;/math&gt; such that &lt;math&gt;\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}&lt;/math&gt;. Then the area enclosed by &lt;math&gt;P&lt;/math&gt; can be written in the form &lt;math&gt;n\sqrt{3}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is an integer. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. ([[2014 AIME II Problems/Problem 10|Solution]])<br /> * Problem 12 - 15: '''5.5'''<br /> *: ''Let<br /> <br /> &lt;cmath&gt;P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).&lt;/cmath&gt;<br /> Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;i = \sqrt { - 1},&lt;/math&gt; and &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> <br /> &lt;cmath&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/cmath&gt;<br /> where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;.'' ([[2003 AIME II Problems/Problem 15|Solution]])<br /> <br /> === [[ARML]] ===<br /> <br /> * Individuals, Problem 1: '''2'''<br /> <br /> * Individuals, Problems 3, 5, 7, and 9: '''3'''<br /> <br /> * Individuals, Problems 2 and 4: '''3'''<br /> <br /> * Individuals, Problems 6 and 8: '''4''' <br /> <br /> * Individuals, Problem 10: '''6'''<br /> <br /> * Team/power, Problem 1-5: '''3.5''' <br /> <br /> * Team/power, Problem 6-10: '''5'''<br /> <br /> ===[[HMMT|HMMT (November)]]===<br /> * Individual Round, Problem 6-8: '''4'''<br /> * Individual Round, Problem 10: '''4.5'''<br /> * Team Round: '''5'''<br /> * Guts: '''3.5-5.25'''<br /> <br /> ===[[CEMC|CEMC Euclid]]===<br /> <br /> * Problem 1-6: '''1-3'''<br /> * Problem 7-10: '''3-6'''<br /> <br /> ===[[Purple Comet! Math Meet|Purple Comet]]===<br /> <br /> * Problems 1-10 (MS): '''2-3'''<br /> * Problems 11-20 (MS): '''3-5'''<br /> * Problems 1-10 (HS): '''2-4'''<br /> * Problems 11-20 (HS): '''4-5'''<br /> * Problems 21-30 (HS): '''5-6'''<br /> <br /> === [[Philippine Mathematical Olympiad Qualifying Round]] ===<br /> <br /> * Problem 1-15: '''2'''<br /> * Problem 16-25: '''3'''<br /> * Problem 26-30: '''4'''<br /> <br /> ==Beginner Olympiad Competitions==<br /> This category consists of beginning Olympiad math competitions. Most junior and first stage Olympiads fall under this category. The range from the difficulty scale would be around 4 to 6. A full list is available [http://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3ABeginner+Olympiad+mathematics+competitions here].<br /> <br /> === [[USAMTS]] ===<br /> USAMTS generally has a different feel to it than olympiads, and is mainly for proofwriting practice instead of olympiad practice depending on how one takes the test. USAMTS allows an entire month to solve problems, with internet resources and books being allowed. However, the ultimate gap is that it permits computer programs to be used, and that Problem 1 is not a proof problem. However, it can still be roughly put to this rating scale:<br /> * Problem 1-2: '''3-4'''<br /> *: ''Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter.'' ([http://usamts.org/Solutions/Solution2_3_16.pdf Solution])<br /> * Problem 3-5: '''5-6'''<br /> *: ''Call a positive real number groovy if it can be written in the form &lt;math&gt;\sqrt{n} + \sqrt{n + 1}&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;. Show that if &lt;math&gt;x&lt;/math&gt; is groovy, then for any positive integer &lt;math&gt;r&lt;/math&gt;, the number &lt;math&gt;x^r&lt;/math&gt; is groovy as well.'' ([http://usamts.org/Solutions/Solutions_20_1.pdf Solution])<br /> <br /> === [[Indonesia Mathematical Olympiad|Indonesia MO]] ===<br /> * Problem 1/5: '''3.5'''<br /> *: '' In a drawer, there are at most &lt;math&gt;2009&lt;/math&gt; balls, some of them are white, the rest are blue, which are randomly distributed. If two balls were taken at the same time, then the probability that the balls are both blue or both white is &lt;math&gt;\frac12&lt;/math&gt;. Determine the maximum amount of white balls in the drawer, such that the probability statement is true?'' &lt;url&gt;viewtopic.php?t=294065 (Solution)&lt;/url&gt;<br /> * Problem 2/6: '''4.5'''<br /> *: ''Find the lowest possible values from the function<br /> &lt;math&gt;f(x) = x^{2008} - 2x^{2007} + 3x^{2006} - 4x^{2005} + 5x^{2004} - \cdots - 2006x^3 + 2007x^2 - 2008x + 2009&lt;/math&gt;<br /> <br /> for any real numbers &lt;math&gt;x&lt;/math&gt;.''&lt;url&gt;viewtopic.php?t=294067 (Solution)&lt;/url&gt;<br /> * Problem 3/7: '''5'''<br /> *: ''A pair of integers &lt;math&gt;(m,n)&lt;/math&gt; is called ''good'' if<br /> &lt;math&gt;m\mid n^2 + n \ \text{and} \ n\mid m^2 + m&lt;/math&gt;<br /> <br /> Given 2 positive integers &lt;math&gt;a,b &gt; 1&lt;/math&gt; which are relatively prime, prove that there exists a ''good'' pair &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;a\mid m&lt;/math&gt; and &lt;math&gt;b\mid n&lt;/math&gt;, but &lt;math&gt;a\nmid n&lt;/math&gt; and &lt;math&gt;b\nmid m&lt;/math&gt;.'' &lt;url&gt;viewtopic.php?t=294068 (Solution)&lt;/url&gt;<br /> * Problem 4/8: '''6'''<br /> *: ''Given an acute triangle &lt;math&gt;ABC&lt;/math&gt;. The incircle of triangle &lt;math&gt;ABC&lt;/math&gt; touches &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively at &lt;math&gt;D,E,F&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle A&lt;/math&gt; cuts &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;DF&lt;/math&gt; respectively at &lt;math&gt;K&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt;. Suppose &lt;math&gt;AA_1&lt;/math&gt; is one of the altitudes of triangle &lt;math&gt;ABC&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> (a) Prove that &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CL&lt;/math&gt; are perpendicular with the angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt;.<br /> <br /> (b) Show that &lt;math&gt;A_1KML&lt;/math&gt; is a cyclic quadrilateral.'' &lt;url&gt;viewtopic.php?t=294069 (Solution)&lt;/url&gt;<br /> <br /> === [[Central American Olympiad]] ===<br /> * Problem 1: '''4'''<br /> *: ''Find all three-digit numbers &lt;math&gt;abc&lt;/math&gt; (with &lt;math&gt;a \neq 0&lt;/math&gt;) such that &lt;math&gt;a^{2} + b^{2} + c^{2}&lt;/math&gt; is a divisor of 26.'' (&lt;url&gt;viewtopic.php?p=903856#903856 Solution&lt;/url&gt;)<br /> * Problem 2,4,5: '''5-6'''<br /> *: ''Show that the equation &lt;math&gt;a^{2}b^{2} + b^{2}c^{2} + 3b^{2} - c^{2} - a^{2} = 2005&lt;/math&gt; has no integer solutions.'' (&lt;url&gt;viewtopic.php?p=291301#291301 Solution&lt;/url&gt;)<br /> * Problem 3/6: '''6.5''' <br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral. &lt;math&gt;I = AC\cap BD&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are points on &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CD&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt; respectively, such that &lt;math&gt;EF \cap GH = I&lt;/math&gt;. If &lt;math&gt;M = EG \cap AC&lt;/math&gt;, &lt;math&gt;N = HF \cap AC&lt;/math&gt;, show that &lt;math&gt;\frac {AM}{IM}\cdot \frac {IN}{CN} = \frac {IA}{IC}&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=828841#p828841 Solution&lt;/url&gt;<br /> <br /> === [[JBMO]] ===<br /> <br /> * Problem 1: '''4'''<br /> *: ''Find all real numbers &lt;math&gt;a,b,c,d&lt;/math&gt; such that <br /> &lt;cmath&gt; \left\{\begin{array}{cc}a+b+c+d = 20,\\ ab+ac+ad+bc+bd+cd = 150.\end{array}\right. &lt;/cmath&gt;''<br /> * Problem 2: '''5'''<br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral with &lt;math&gt;\angle DAC=\angle BDC=36^\circ&lt;/math&gt;, &lt;math&gt;\angle CBD=18^\circ&lt;/math&gt; and &lt;math&gt;\angle BAC=72^\circ&lt;/math&gt;. The diagonals intersect at point &lt;math&gt;P&lt;/math&gt;. Determine the measure of &lt;math&gt;\angle APD&lt;/math&gt;.''<br /> * Problem 3: '''5'''<br /> *: ''Find all prime numbers &lt;math&gt;p,q,r&lt;/math&gt;, such that &lt;math&gt;\frac pq-\frac4{r+1}=1&lt;/math&gt;.''<br /> * Problem 4: '''6'''<br /> *: ''A &lt;math&gt;4\times4&lt;/math&gt; table is divided into &lt;math&gt;16&lt;/math&gt; white unit square cells. Two cells are called neighbors if they share a common side. A '''move''' consists in choosing a cell and changing the colors of neighbors from white to black or from black to white. After exactly &lt;math&gt;n&lt;/math&gt; moves all the &lt;math&gt;16&lt;/math&gt; cells were black. Find all possible values of &lt;math&gt;n&lt;/math&gt;.''<br /> <br /> ==Olympiad Competitions==<br /> This category consists of standard Olympiad competitions, usually ones from national Olympiads. Average difficulty is from 5.5 to 8. A full list is available [http://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AOlympiad+mathematics+competitions here].<br /> <br /> === [[USAJMO]] ===<br /> * Problem 1/4: '''5.75'''<br /> * Problem 2/5: '''6.5'''<br /> * Problem 3/6: '''7.25'''<br /> <br /> ===[[HMMT|HMMT (February)]]===<br /> * Individual Round, Problem 1-5: '''5'''<br /> * Individual Round, Problem 6-10: '''6.5'''<br /> * Team Round: '''7.5'''<br /> * HMIC: '''8'''<br /> <br /> === [[Canadian MO]] ===<br /> <br /> * Problem 1: '''5.5'''<br /> * Problem 2: '''6'''<br /> * Problem 3: '''6.5''' <br /> * Problem 4: '''7-7.5'''<br /> * Problem 5: '''7.5-8'''<br /> <br /> === Austrian MO ===<br /> <br /> * Regional Competition for Advanced Students, Problems 1-4: '''5''' <br /> * Federal Competition for Advanced Students, Part 1. Problems 1-4: '''6''' <br /> * Federal Competition for Advanced Students, Part 2, Problems 1-6: '''7'''<br /> <br /> === [[Ibero American Olympiad]] ===<br /> <br /> * Problem 1/4: '''5.5'''<br /> * Problem 2/5: '''6.5'''<br /> * Problem 3/6: '''7.5'''<br /> <br /> === [[APMO]] ===<br /> *Problem 1: '''6'''<br /> *Problem 2: '''7'''<br /> *Problem 3: '''7'''<br /> *Problem 4: '''7.5'''<br /> *Problem 5: '''8.5'''<br /> <br /> === Balkan MO ===<br /> <br /> * Problem 1: '''6'''<br /> *: '' Solve the equation &lt;math&gt;3^x - 5^y = z^2&lt;/math&gt; in positive integers. '' <br /> * Problem 2: '''6.5'''<br /> *: '' Let &lt;math&gt;MN&lt;/math&gt; be a line parallel to the side &lt;math&gt;BC&lt;/math&gt; of a triangle &lt;math&gt;ABC&lt;/math&gt;, with &lt;math&gt;M&lt;/math&gt; on the side &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; on the side &lt;math&gt;AC&lt;/math&gt;. The lines &lt;math&gt;BN&lt;/math&gt; and &lt;math&gt;CM&lt;/math&gt; meet at point &lt;math&gt;P&lt;/math&gt;. The circumcircles of triangles &lt;math&gt;BMP&lt;/math&gt; and &lt;math&gt;CNP&lt;/math&gt; meet at two distinct points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Prove that &lt;math&gt;\angle BAQ = \angle CAP&lt;/math&gt;. ''<br /> * Problem 3: '''7.5'''<br /> *: '' A &lt;math&gt;9 \times 12&lt;/math&gt; rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres &lt;math&gt;C_1,C_2...,C_{96}&lt;/math&gt; in such way that the following to conditions are both fulfilled<br /> <br /> &lt;math&gt;(i)&lt;/math&gt; the distances &lt;math&gt;C_1C_2,...C_{95}C_{96}, C_{96}C_{1}&lt;/math&gt; are all equal to &lt;math&gt;\sqrt {13}&lt;/math&gt;<br /> <br /> &lt;math&gt;(ii)&lt;/math&gt; the closed broken line &lt;math&gt;C_1C_2...C_{96}C_1&lt;/math&gt; has a centre of symmetry? ''<br /> * Problem 4: '''8'''<br /> *: '' Denote by &lt;math&gt;S&lt;/math&gt; the set of all positive integers. Find all functions &lt;math&gt;f: S \rightarrow S&lt;/math&gt; such that<br /> <br /> &lt;math&gt;f \bigg(f^2(m) + 2f^2(n)\bigg) = m^2 + 2 n^2&lt;/math&gt; for all &lt;math&gt;m,n \in S&lt;/math&gt;. '<br /> <br /> ==Hard Olympiad Competitions==<br /> This category consists of harder Olympiad contests. Difficulty is usually from 7 to 10. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AHard+Olympiad+mathematics+competitions here].<br /> <br /> === [[USAMO]] ===<br /> * Problem 1/4: '''7'''<br /> *: ''Let &lt;math&gt;\mathcal{P}&lt;/math&gt; be a convex polygon with &lt;math&gt;n&lt;/math&gt; sides, &lt;math&gt;n\ge3&lt;/math&gt;. Any set of &lt;math&gt;n - 3&lt;/math&gt; diagonals of &lt;math&gt;\mathcal{P}&lt;/math&gt; that do not intersect in the interior of the polygon determine a ''triangulation'' of &lt;math&gt;\mathcal{P}&lt;/math&gt; into &lt;math&gt;n - 2&lt;/math&gt; triangles. If &lt;math&gt;\mathcal{P}&lt;/math&gt; is regular and there is a triangulation of &lt;math&gt;\mathcal{P}&lt;/math&gt; consisting of only isosceles triangles, find all the possible values of &lt;math&gt;n&lt;/math&gt;.'' ([[2008 USAMO Problems/Problem 4|Solution]]) <br /> * Problem 2/5: '''8'''<br /> *: ''Three nonnegative real numbers &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, &lt;math&gt;r_3&lt;/math&gt; are written on a blackboard. These numbers have the property that there exist integers &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, not all zero, satisfying &lt;math&gt;a_1r_1 + a_2r_2 + a_3r_3 = 0&lt;/math&gt;. We are permitted to perform the following operation: find two numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; on the blackboard with &lt;math&gt;x \le y&lt;/math&gt;, then erase &lt;math&gt;y&lt;/math&gt; and write &lt;math&gt;y - x&lt;/math&gt; in its place. Prove that after a finite number of such operations, we can end up with at least one &lt;math&gt;0&lt;/math&gt; on the blackboard.'' ([[2008 USAMO Problems/Problem 5|Solution]])<br /> * Problem 3/6: '''9'''<br /> *: ''Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree &lt;math&gt;n &lt;/math&gt; with real coefficients is the average of two monic polynomials of degree &lt;math&gt;n &lt;/math&gt; with &lt;math&gt;n &lt;/math&gt; real roots.'' ([[2002 USAMO Problems/Problem 3|Solution]])<br /> <br /> === [[USA TST]] ===<br /> <br /> (seems to vary more than other contests; estimates based on 08 and 09)<br /> <br /> * Problem 1/4/7: '''7'''<br /> * Problem 2/5/8: '''8'''<br /> * Problem 3/6/9: '''9.5'''<br /> <br /> === [[Putnam]] ===<br /> <br /> * Problem A/B,1-2: '''7'''<br /> *: ''Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola &lt;math&gt;xy = 1&lt;/math&gt; and both branches of the hyperbola &lt;math&gt;xy = - 1.&lt;/math&gt; (A set &lt;math&gt;S&lt;/math&gt; in the plane is called ''convex'' if for any two points in &lt;math&gt;S&lt;/math&gt; the line segment connecting them is contained in &lt;math&gt;S.&lt;/math&gt;)'' ([https://artofproblemsolving.com/community/c7h177227p978383 Solution])<br /> * Problem A/B,3-4: '''8'''<br /> *: ''Let &lt;math&gt;H&lt;/math&gt; be an &lt;math&gt;n\times n&lt;/math&gt; matrix all of whose entries are &lt;math&gt;\pm1&lt;/math&gt; and whose rows are mutually orthogonal. Suppose &lt;math&gt;H&lt;/math&gt; has an &lt;math&gt;a\times b&lt;/math&gt; submatrix whose entries are all &lt;math&gt;1.&lt;/math&gt; Show that &lt;math&gt;ab\le n&lt;/math&gt;.'' ([https://artofproblemsolving.com/community/c7h64435p383280 Solution])<br /> * Problem A/B,5-6: '''9'''<br /> *: ''For any &lt;math&gt;a &gt; 0&lt;/math&gt;, define the set &lt;math&gt;S(a) = \{[an]|n = 1,2,3,...\}&lt;/math&gt;. Show that there are no three positive reals &lt;math&gt;a,b,c&lt;/math&gt; such that &lt;math&gt;S(a)\cap S(b) = S(b)\cap S(c) = S(c)\cap S(a) = \emptyset,S(a)\cup S(b)\cup S(c) = \{1,2,3,...\}&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?t=127810 Solution&lt;/url&gt;)<br /> <br /> === [[China TST]] ===<br /> <br /> * Problem 1/4: '''7''' <br /> *: ''Given an integer &lt;math&gt;m,&lt;/math&gt; prove that there exist odd integers &lt;math&gt;a,b&lt;/math&gt; and a positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;cmath&gt;2m=a^{19}+b^{99}+k*2^{1000}.&lt;/cmath&gt;''<br /> * Problem 2/5: '''8.5''' <br /> *: ''Given a positive integer &lt;math&gt;n&gt;1&lt;/math&gt; and real numbers &lt;math&gt;a_1 &lt; a_2 &lt; \ldots &lt; a_n,&lt;/math&gt; such that &lt;math&gt;\dfrac{1}{a_1} + \dfrac{1}{a_2} + \ldots + \dfrac{1}{a_n} \le 1,&lt;/math&gt; prove that for any positive real number &lt;math&gt;x,&lt;/math&gt; &lt;cmath&gt;\left(\dfrac{1}{a_1^2+x} + \dfrac{1}{a_2^2+x} + \ldots + \dfrac{1}{a_n^2+x}\right)^2 \ge \dfrac{1}{2a_1(a_1-1)+2x}.&lt;/cmath&gt;''<br /> * Problem 3/6: '''10'''<br /> *: ''Let &lt;math&gt;n&gt;1&lt;/math&gt; be an integer and let &lt;math&gt;a_0,a_1,\ldots,a_n&lt;/math&gt; be non-negative real numbers. Define &lt;math&gt;S_k=\sum_{i=0}^k \binom{k}{i}a_i&lt;/math&gt; for &lt;math&gt;k=0,1,\ldots,n&lt;/math&gt;. Prove that&lt;cmath&gt;\frac{1}{n} \sum_{k=0}^{n-1} S_k^2-\frac{1}{n^2}\left(\sum_{k=0}^{n} S_k\right)^2\le \frac{4}{45} (S_n-S_0)^2.&lt;/cmath&gt;''<br /> <br /> === [[IMO]] ===<br /> <br /> * Problem 1/4: '''6.5'''<br /> *: ''Find all functions &lt;math&gt;f: (0, \infty) \mapsto (0, \infty)&lt;/math&gt; (so that &lt;math&gt;f&lt;/math&gt; is a function from the positive real numbers) such that<br /> &lt;center&gt;&lt;math&gt;\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}&lt;/math&gt;&lt;/center&gt; for all positive real numbers &lt;math&gt;w,x,y,z,&lt;/math&gt; satisfying &lt;math&gt;wx = yz.&lt;/math&gt; ([[2008 IMO Problems/Problem 4|Solution]])<br /> ''<br /> * Problem 2/5: '''7.5-8'''<br /> *: ''Let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial of degree &lt;math&gt;n&gt;1&lt;/math&gt; with integer coefficients, and let &lt;math&gt;k&lt;/math&gt; be a positive integer. Consider the polynomial &lt;math&gt;Q(x) = P( P ( \ldots P(P(x)) \ldots ))&lt;/math&gt;, where &lt;math&gt;P&lt;/math&gt; occurs &lt;math&gt;k&lt;/math&gt; times. Prove that there are at most &lt;math&gt;n&lt;/math&gt; integers &lt;math&gt;t&lt;/math&gt; such that &lt;math&gt;Q(t)=t&lt;/math&gt;.'' ([[2006 IMO Problems/Problem 5|Solution]])<br /> * Problem 3/6: '''9-10'''<br /> *: ''Assign to each side &lt;math&gt;b&lt;/math&gt; of a convex polygon &lt;math&gt;P&lt;/math&gt; the maximum area of a triangle that has &lt;math&gt;b&lt;/math&gt; as a side and is contained in &lt;math&gt;P&lt;/math&gt;. Show that the sum of the areas assigned to the sides of &lt;math&gt;P&lt;/math&gt; is at least twice the area of &lt;math&gt;P&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=572824#572824 Solution&lt;/url&gt;)<br /> <br /> === [[IMO Shortlist]] ===<br /> <br /> * Problem 1-2: '''5.5-7'''<br /> * Problem 3-4: '''7-8'''<br /> * Problem 5+: '''8-10'''<br /> <br /> [[Category:Mathematics competitions]]</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=AMC_8&diff=105050 AMC 8 2019-03-30T17:36:58Z <p>Zhang2018: /* Format */</p> <hr /> <div>The '''AMC 8''' is an exam for students in grades 8 and below, administered annually by the [[American Mathematics Competitions]] (AMC) to students all over the United States.<br /> <br /> Usually, high scoring students will be given a chance by their school to take the more challenging [[AMC 10]] exam. However, there is no requirement for the AMC 10 besides the fact that you have to be 10&lt;math&gt;^\text{th}&lt;/math&gt; grade or below. <br /> <br /> The AMC 8 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC. The test is intended to foster interest in mathematics and also to help middle school students learn [[mathematical problem solving]].<br /> <br /> {{Contest Info|name=AMC 8|region=USA|type=Multiple Choice|difficulty=1 - 1.5|breakdown=&lt;u&gt;Problems 1 - 12&lt;/u&gt;: 1&lt;br&gt;&lt;u&gt;Problems 13 - 25&lt;/u&gt;: 1.5}}<br /> <br /> == Format ==<br /> <br /> The AMC 8 Exam is a 25 problem exam. There are 40 minutes given in exam. Problems increase in difficulty as the problem number increases.<br /> <br /> ==Schedule==<br /> <br /> The AMC 8 is usually administered on the third Tuesday of November. Some schools may take it as late as the fourth Tuesday. The 2017 AMC 8 was administered on November 14. 2018 was on the second tuesday.<br /> <br /> == Curriculum ==<br /> The AMC 8 tests [[mathematical problem solving]] with [[algebra]], [[arithmetic]], [[counting]], [[geometry]], [[logic]], [[number theory]], and [[probability]].<br /> <br /> == Resources ==<br /> === Links ===<br /> * [[AMC 8 Problems and Solutions]]<br /> Use the links above to access many years worth of competition questions.<br /> <br /> === Recommended reading ===<br /> * Introduction to Counting &amp; Probability by Dr. [[David Patrick]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3 Information]<br /> * Introduction to Geometry by [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9 Information]<br /> * The Art of Problem Solving Volume I by [[Sandor Lehoczky]] and [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=1 Information].<br /> <br /> <br /> === Preparation Classes ===<br /> * [[Art of Problem Solving]] offers many [http://www.artofproblemsolving.com/Classes/AoPS_C_About.php helpful online classes] on topics covered by the AMC 8.<br /> * [[AoPS]] holds many free [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jams], some of which include problems and concepts at the level tested by the AMC 8. <br /> * [[EPGY]] offers some contest preparation classes.<br /> <br /> == See also ==<br /> * [[Mathematics competitions]]<br /> * [[MathCounts]]<br /> * [[MOEMS]]<br /> [[Category:Mathematics competitions]]<br /> <br /> [[Category:Introductory mathematics competitions]]</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=AMC_8&diff=105049 AMC 8 2019-03-30T17:36:46Z <p>Zhang2018: /* Format */</p> <hr /> <div>The '''AMC 8''' is an exam for students in grades 8 and below, administered annually by the [[American Mathematics Competitions]] (AMC) to students all over the United States.<br /> <br /> Usually, high scoring students will be given a chance by their school to take the more challenging [[AMC 10]] exam. However, there is no requirement for the AMC 10 besides the fact that you have to be 10&lt;math&gt;^\text{th}&lt;/math&gt; grade or below. <br /> <br /> The AMC 8 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC. The test is intended to foster interest in mathematics and also to help middle school students learn [[mathematical problem solving]].<br /> <br /> {{Contest Info|name=AMC 8|region=USA|type=Multiple Choice|difficulty=1 - 1.5|breakdown=&lt;u&gt;Problems 1 - 12&lt;/u&gt;: 1&lt;br&gt;&lt;u&gt;Problems 13 - 25&lt;/u&gt;: 1.5}}<br /> <br /> == Format ==<br /> <br /> The AMC 8 Exam is a 25 problem exam. There are 40 minutes given in exam. Problem increase in difficulty as the problem number increases.<br /> <br /> ==Schedule==<br /> <br /> The AMC 8 is usually administered on the third Tuesday of November. Some schools may take it as late as the fourth Tuesday. The 2017 AMC 8 was administered on November 14. 2018 was on the second tuesday.<br /> <br /> == Curriculum ==<br /> The AMC 8 tests [[mathematical problem solving]] with [[algebra]], [[arithmetic]], [[counting]], [[geometry]], [[logic]], [[number theory]], and [[probability]].<br /> <br /> == Resources ==<br /> === Links ===<br /> * [[AMC 8 Problems and Solutions]]<br /> Use the links above to access many years worth of competition questions.<br /> <br /> === Recommended reading ===<br /> * Introduction to Counting &amp; Probability by Dr. [[David Patrick]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3 Information]<br /> * Introduction to Geometry by [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9 Information]<br /> * The Art of Problem Solving Volume I by [[Sandor Lehoczky]] and [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=1 Information].<br /> <br /> <br /> === Preparation Classes ===<br /> * [[Art of Problem Solving]] offers many [http://www.artofproblemsolving.com/Classes/AoPS_C_About.php helpful online classes] on topics covered by the AMC 8.<br /> * [[AoPS]] holds many free [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jams], some of which include problems and concepts at the level tested by the AMC 8. <br /> * [[EPGY]] offers some contest preparation classes.<br /> <br /> == See also ==<br /> * [[Mathematics competitions]]<br /> * [[MathCounts]]<br /> * [[MOEMS]]<br /> [[Category:Mathematics competitions]]<br /> <br /> [[Category:Introductory mathematics competitions]]</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=104899 AoPS Wiki:FAQ 2019-03-24T15:05:42Z <p>Zhang2018: /* Can I change my username? */</p> <hr /> <div>{{shortcut|[[A:FAQ]]}}<br /> <br /> This is a community created list of Frequently Asked Questions about Art of Problem Solving. If you have a request to edit or add a question on this page, please make it [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=416&amp;t=414129 here].<br /> <br /> == General==<br /> <br /> <br /> ==== Can I change my username? ====<br /> <br /> No, unless you contact Art of Problem Solving<br /> <br /> ==== Why can't I change my avatar? ====<br /> <br /> : You must be a user for two weeks before being able to change your avatar.<br /> : You can get another avatar by downloading a picture you like and changing the dimensions via Paint or Paint3D; however, it's really hard to get a good-looking picture under the size limit.<br /> : Just be happy with what you have.<br /> <br /> ==== Something looks weird (e.g., blurry, missing line, etc.) ====<br /> <br /> : This is likely due to your browser zoom level. Please make sure your zoom level is at 100% for correct rendering of the web page.<br /> <br /> ==== Can I make more than one account?==== <br /> <br /> :Short answer: No.<br /> <br /> :Long answer: Multiple accounts (multis) are banned on AoPS. Having more than one account leads to issues of not remembering on what account you did what. Using multiple accounts to &quot;game&quot; the system, (e.g. increase rating for posts or in online games) will lead to bans on all accounts associated with you. If you have already made additional accounts, please choose one account and stop using the others.<br /> <br /> ====What software does Art of Problem Solving use to run the website?====<br /> <br /> :* Search: Solr<br /> :* Wiki: MediaWiki<br /> :* Asymptote and &lt;math&gt;\text{\LaTeX}&lt;/math&gt; are generated through their respective binary packages<br /> :* Videos: YouTube<br /> <br /> :All other parts of the website are custom built.<br /> <br /> ====Can you make an AoPS App?====<br /> <br /> :No. Contrary to popular opinion, any app built would be less functional than the website. Any features missing on the website in phone view would also be missing in an app as an app doesn't magically make the screen bigger. In addition, the logistics and costs involved are too great.<br /> <br /> == Forums ==<br /> <br /> ====How do I create a forum?====<br /> <br /> :To create an AoPS forum, a user must be on the AoPS community for at least 2 weeks. To create a forum, hover over the community tab, then click &quot;My AoPS.&quot; You should now see your avatar, and a list of your friends. Now, click on the &quot;My Forums&quot; tab. There, you would be able to see which forum you moderate or administrate, as well as the private forums you can access. Click on the &quot;+&quot; button at the top right. This should lead you to a forum creating page.<br /> <br /> ====Why do some posts say they were posted in the future?====<br /> :The AoPS clock is based on the official Naval Observatory time. This is considered the most accurate time. If your computer's system clock is behind the correct time, recent posts may indicate they were posted in the future. Please correct your computer's clock or enable clock synchronization so that your clock is always correct. Mac users may wish to check [http://www.macinstruct.com/node/92 Synchronize your Mac's Clock With A Time Server]. You can also check [http://www.time.gov the US offical time.]<br /> <br /> ====I've lost admin access to a forum or blog I created, how do I get it back?====<br /> :Please send an email to sheriff@aops.com. They will research your request and restore admin access to your forum if appropriate.<br /> <br /> ====What should I do if I find a glitch in the community?====<br /> :First, search the Site Support forum to check if your issue has already been reported. You can search here.<br /> <br /> :If your issue isn't reported, try refreshing your browser page. Most issues go away after a refresh and there is no need to report the issue unless it continues after you refresh your browser. You can refresh on most browsers in Windows with Ctrl + R and on Mac with Cmd + R. To Hard Refresh, click Shift between Ctrl/Cmd and R.<br /> <br /> :Some commonly known glitches that should not be reported are avatars appearing twice in the topic list or private message, friends appearing more than once in the friend list, the edit icon showing next to a message when it shouldn't, and not being able to search for a forum after going back a page in the community.<br /> <br /> ==== How do I format my post, e.g. bold text, add URLs, etc.? ====<br /> <br /> :AoPS is based on a markup language called BBCode. A tutorial of its functions on AoPS and how to use them can be found [[BBCode:Tutorial|here]].<br /> <br /> ==== How do I hide content in the forums? ====<br /> :Wrap the content you want to hide in [hide] tags.<br /> [hide]Content[/hide]<br /> :If you want to customize the label, instead of saying &quot;Click here to reveal hidden text&quot;, you can do something like:<br /> [hide=Label to display]Content[/hide]<br /> <br /> ====I got the message &quot;Too many messages.&quot; when trying to send a private message, why?====<br /> :To prevent PM spam abuse like &quot;hvbowibvibviorybvirusbshbiuovvisuvib&quot;, users with less than five forum posts are limited to four private messages within a forty-eight hour period.<br /> <br /> ==== If I make more posts, it means I'm a better user, right? ====<br /> :Post quality is far more important than post quantity.<br /> <br /> ==== I have made some posts but my post count did not increase. Why? ====<br /> :When you post in some of the forums, such as the Test Forum, Mafia Forum, Fun Factory, and most user created forums, the post does not count towards your overall post count. And still. Quality over quantity.<br /> <br /> ==== I believe a post needs corrective action. What should I do? ====<br /> :If you believe a post needs moderative action, you may report it by clicking the [aops]z[/aops] icon on the upper-right corner of that post. If it's a minor mistake, you may want to PM the offending user instead and explain how they can make their post better. Usually, you shouldn't publicly post such things on a thread itself, which is called &quot;backseat moderation&quot; and is considered rude and unproductive.<br /> <br /> ==== How do I post images? ====<br /> :While AoPS forums have the ability to attach images, we do not generally recommend doing so, as we can not guarantee the images will be available through upgrades, restorations, etc. We also have limited disk space which causes us to remove attachments from time to time. Therefore, we recommend using a third party image hosting solution. There are many options, but we recommend imgur.com.<br /> <br /> :* Go to imgur.com<br /> :* Click New Post<br /> :* Upload your image<br /> :* Hover over image. A URL, Copy button, and down arrow will appear<br /> :* Click the down arrow<br /> :* Click Get Share Links<br /> :* Click Copy next to the BBCode (Forums) option<br /> :* Paste the coped BBCode into AoPS<br /> <br /> ====How do I become a moderator of a forum?====<br /> :The creators of user-created forums select moderators in several different ways. Contact the creator(s) of the forum for more information.<br /> <br /> :The AoPS staff will select moderators for the major forums on an as-needed basis. If the AoPS staff determines that a forum needs additional moderators, then the staff will reach out to productive users and invite them to become moderators. These are some features that the AoPS staff may consider when selecting new moderators.<br /> <br /> :* An established history of productive posting. No unproductive, spammy, or troll posts in the past several years. (For example, a high school student might have made some poor choices while in elementary school -- but then learned from those mistakes.)<br /> :* No use of \8char or other strategies to get around rules or restrictions on the forums.<br /> :* When the user reports posts, the report is useful and helpful, and the reported post (or thread, or user) really did reflect a serious issue worthy of administrators' attention. The user filled in the &quot;Further details&quot; field with valuable information. The user does not report the use of phrases such as &quot;gosh, golly gee&quot; as profanity.<br /> :* No multis.<br /> :* No backseat moderating, no posting of &quot;\requestlock&quot;.<br /> :* When the user revives an old thread, the new post is valuable and relevant.<br /> :* The user does not use the forums to cheat on Alcumus, homework, contests, or similar activities.<br /> <br /> == Blogs ==<br /> ==== How come I can't create a blog? ====<br /> :One needs to have at least 5 posts in order to make a blog.<br /> <br /> ==== How do I make my blog look nice? ====<br /> :Many AoPSers make their blogs look awesome by applying [[CSS|CSS]], which is a high-level stylesheet language. This can be done by typing CSS code into the CSS box in the Blog Control Panel.<br /> <br /> == Alcumus ==<br /> <br /> ==== How is rating computed? ====<br /> :The rating is more of a prediction of what percentage of problems in the topic the Alcumus engine believes you will get correct. As you get more and more correct, the rating will go up slower and slower. However, if you are predicted to get most correct, and you miss one or two problems, the rating, or prediction of percentage correct, will go down.<br /> <br /> ==== I am stuck on a problem, and changing the topic does not change the problem. ====<br /> :Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before moving on.<br /> <br /> ==== Why can't I change topics? ====<br /> :Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before the topic changes to the currently selected topic.<br /> <br /> == Contests ==<br /> ==== Where can I find past contest questions and solutions? ====<br /> :In the [http://www.artofproblemsolving.com/Forum/resources.php Contests] section.<br /> <br /> ==== How do I get problems onto the contest page? ====<br /> <br /> :Make a topic for each question in the appropriate forum, copy/paste the urls to the National Olympiad. Your problems may eventually be submitted into the Contest page.<br /> <br /> ==== What are the guidelines for posting problems to be added to the contests section? ====<br /> :Refer to the [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=144&amp;t=195579 guidelines in this post].<br /> <br /> ==== Why is the wiki missing many contest questions? ====<br /> :Generally, it is because users have not yet posted them onto the wiki (translation difficulties, not having access to the actual problems, lack of interest, etc). If you have a copy, please post the problems in the Community Section! In some cases, however, problems may be missing due to copyright claims from maths organizations.<br /> <br /> ====Where are all of the mock contests?====<br /> :User-generated mock contests are now in its separate subforum, [https://artofproblemsolving.com/community/c594864_aops_mock_contests AoPS Mock Contests]. Please use that forum for creating, hosting, participating in, or recruiting writers/test solvers for mock contests.<br /> <br /> ====How can I host a mock contest?====<br /> :See posts on [https://artofproblemsolving.com/community/c594864h1168324_hi__im_interested_in_writing_a_mock_contest writing a mock contest] and [https://artofproblemsolving.com/community/c594864h1349281_tips_for_organizing_a_mock_contest Tips for Organizing a Mock Contest].<br /> <br /> == LaTeX and Asymptote ==<br /> ==== What is LaTeX, and how do I use it? ====<br /> <br /> :&lt;math&gt;\text{\LaTeX}&lt;/math&gt; is a typesetting markup language and document preparation system. It is widely used for typesetting expressions containing mathematical formulae. See [https://artofproblemsolving.com/wiki/index.php/LaTeX:About LaTeX:About].<br /> <br /> ==== How can I download LaTeX to use on the forums? ====<br /> <br /> :There are no downloads necessary; the forums and the wiki render LaTeX commands between dollar signs. <br /> <br /> ==== How can I download LaTeX for personal use? ====<br /> :You can download TeXstudio [http://texstudio.sourceforge.net here] or TeXnicCenter [http://www.texniccenter.org here]<br /> <br /> ==== Where can I find a list of LaTeX commands? ====<br /> :See [[LaTeX:Symbols|here]].<br /> <br /> ==== Where can I test LaTeX commands? ====<br /> <br /> :[[A:SAND|Sandbox]] or [http://www.artofproblemsolving.com/Resources/texer.php TeXeR]. You can also use our [http://artofproblemsolving.com/community/c67_test_forum Test Forum].<br /> <br /> ==== Where can I find examples of Asymptote diagrams and code? ====<br /> <br /> :Search this wiki for the &lt;tt&gt;&lt;nowiki&gt;&lt;asy&gt;&lt;/nowiki&gt;&lt;/tt&gt; tag or the Forums for the &lt;tt&gt;&lt;nowiki&gt;[asy]&lt;/nowiki&gt;&lt;/tt&gt; tag. See also [[Asymptote:_Useful_commands_and_their_Output|these examples]] and [[Proofs without words|this article]] (click on the images to obtain the code).<br /> <br /> ==== How can I draw 3D diagrams? ====<br /> <br /> :See [[Asymptote: 3D graphics]].<br /> <br /> ==== What is the cse5 package? ==== <br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=149650 here]. The package contains a set of shorthand commands that implement the behavior of usual commands, for example &lt;tt&gt;D()&lt;/tt&gt; for &lt;tt&gt;draw()&lt;/tt&gt; and &lt;tt&gt;dot()&lt;/tt&gt;, and so forth.<br /> <br /> ==== What is the olympiad package? ====<br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=165767 here]. The package contains a set of commands useful for drawing diagrams related to [[:Category:Olympiad Geometry Problems|olympiad geometry problems]].<br /> <br /> == AoPSWiki ==<br /> ==== Is there a guide for wiki syntax? ====<br /> <br /> :See [http://en.wikipedia.org/wiki/Help:Wiki_markup wiki markup], [[AoPSWiki:Tutorial]], and [[Help:Contents]].<br /> <br /> ==== What do I do if I see a mistake in the wiki? ====<br /> <br /> :Click &quot;edit&quot; and correct the error.<br /> <br /> == Miscellaneous ==<br /> ==== Is it possible to join the AoPS Staff? ====<br /> <br /> :Yes. Mr. Rusczyk will sometimes hire a small army of college students to work as interns, graders, and teaching assistants. You must be at least in your second semester of your senior year and be legal to work in the U.S. (at least 16).<br /> <br /> ==== What is the minimum age to be an assistant in an Art of Problem Solving class? ====<br /> <br /> :You must have graduated from high school, or at least be in the second term of your senior year.<br /> <br /> ==AoPS Acronyms==<br /> *'''AFK'''- Away from keyboard, inactive<br /> *'''AoPS'''- Art of Problem Solving<br /> *'''AIME'''- American Invitational Mathematics Examination<br /> *'''admin(s)'''- Administrator(s)<br /> *'''AMC'''- American Math Competitions<br /> *'''AMO'''-USA Mathematical Olympiad<br /> *'''ATM'''- At the Moment/Automated teller machine<br /> *'''brb'''- Be right Back<br /> *'''BTW'''- By the way<br /> *'''CEMC''' - Centre for Mathematics and Computing<br /> *'''Combo''' - Combinatorics (Counting and Probability)<br /> *'''C&amp;P or C+P or CP''' - Counting and Probability or Contests and Programs<br /> *'''cbrt''' - Cube Root<br /> *'''EBWOP'''- Editing by way of post<br /> *'''FF'''- Fun Factory<br /> *'''FTFY''' - Fixed that for you<br /> *'''FTW'''- For the Win, a game on AoPS<br /> *'''gg'''- Good Game<br /> *'''gj'''- Good Job<br /> *'''glhf'''-Good Luck Have Fun<br /> *'''gtg''' - Got to go<br /> *'''HSM''' - High School Math Forum<br /> *'''ID(R or E)K'''-I Don't (Really or Even) Know<br /> *'''ID(R or E)C'''-I Don't (Really or Even) Care<br /> *'''iff'''-If and only if<br /> *'''IIRC'''- If I recall correctly<br /> *'''IKR'''- I know, right?<br /> *'''IMO'''- In my opinion (or International Math Olympiad, depending on context)<br /> *'''IMHO'''- In my humble/honest opinion <br /> *'''JK'''- Just Kidding<br /> *'''JMO'''- United States of America Junior Mathematical Olympiad<br /> *'''lol'''- Laugh Out Loud<br /> *'''MC'''- Mathcounts, a popular math contest for Middle School students.<br /> *'''NFL'''- Not for long/No friends left/National Football League<br /> *'''NHL'''-National Hockey League<br /> *'''mod(s)'''- Moderator(s)<br /> *'''MOEMS'''- Math Olympiads for Elementary and Middle Schools<br /> *'''MO(S)P'''- Mathematical Olympiad (Summer) Program<br /> *'''MSM'''- Middle School Math Forum<br /> *'''NIMO'''-National Internet Math Olympiad<br /> *'''NT'''- Number Theory<br /> *'''OBC'''- Online by computer<br /> *'''OMG'''- Oh My Gosh<br /> *'''OMO'''-Online Math Open<br /> *'''OP'''- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering<br /> *'''QED'''- Quod erat demonstrandum, Latin for &quot;Which was to be proven&quot;; some English mathematicians use it as an acronym for Quite Elegantly Done<br /> *'''QS&amp;A'''- Questions, Suggestions, and Announcements Forum<br /> *'''ro(t)fl''' - Rolling on the floor laughing<br /> *'''smh''' - Shaking my head/somehow<br /> *'''sqrt''' - Square root<br /> *'''Sticky'''- A post pinned to the top of a forum - a thing you really should read<br /> *'''ToS'''- Terms of Service - a thing you really should read<br /> *'''USA(J)MO'''- USA (Junior) Mathematical Olympiadi<br /> *'''V/LA'''- Vacation or Long Absence/Limited Access<br /> *'''WIP'''- Work in Progress<br /> *'''WLOG'''- Without loss of generality<br /> *'''WOOT''' - Worldwide Online Olympiad Training<br /> *'''wrt'''- With respect to<br /> *'''wtg''' - Way to go<br /> *'''tytia'''- Thank you, that is all<br /> *'''ty'''- Thank you<br /> *'''ttyl'''- Talk to you later<br /> *'''xD'''- Bursting Laugh<br /> *'''TL;DR'''- Too long; don't read<br /> *'''yw'''- You're welcome<br /> <br /> == FTW! ==<br /> <br /> Please see the [//artofproblemsolving.com/ftw/faq For the Win! FAQ] for many common questions.<br /> <br /> == School ==<br /> <br /> ==== What if I miss a class? ====<br /> :There are classroom transcripts available under My Classes, available at the top right of the web site. You can view these transcripts in order to review any missed material. You can also ask questions on the class message board. Don't worry, though, classroom participation usually isn't weighted heavily.<br /> <br /> ==== Is there audio or video in class? ====<br /> :There is generally no audio or video in the class. The classes are generally text and image based, in an interactive chat room environment, which allows students to ask questions at any time during the class. In addition to audio and video limiting interactivity and being less pedagogically effective, the technology isn't quite there yet for all students to be able to adequately receive streaming audio and video.<br /> <br /> ==== What if I want to drop out of a class? ====<br /> :For any course with more than 2 classes, students can drop the course any time before the third class begins and receive a full refund. No drops are allowed after the third class has started. To drop the class, go to the My Classes section by clicking the My Classes link at the top-right of the website. Then find the area on the right side of the page that lets you drop the class. A refund will be processed within 10 business days.<br /> <br /> ==== For my homework, there is supposed to be a green bar but it's orange, why? ====<br /> <br /> :For the bar to turn green, the writing problem must be attempted. Once it is attempted the bar will turn green.<br /> <br /> ==== I need more time for my homework, what should I do? ====<br /> <br /> :There is a &quot;Request Extension&quot; button in the homework tab of your class. This will automatically extend the due date to 2 days after the normal deadline. If you want more time you need to ask for it in the little comment box, stating the reason why you want the extension, and how much time you want. This request will be looked at by the teachers and they will decide if you get the extension or not. Note that you can only use this button 3 times. After, you will need administrator approvement.<br /> :Otherwise, you can send an email to extensions@aops.com with your username, class name and ID (the number in the class page URL after https://artofproblemsolving.com/class/) and reason for extension. Someone should get back to you within a couple days.</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11&diff=104633 1983 AIME Problems/Problem 11 2019-03-17T23:32:08Z <p>Zhang2018: /* Solution */</p> <hr /> <div>== Problem ==<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 /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); 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,W);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,SE);<br /> label(&quot;D&quot;,D,NE);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:1983Number11.JPG by bpms --&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> First, we find the height of the solid by dropping a perpendicular from the midpoint of &lt;math&gt;AD&lt;/math&gt; to &lt;math&gt;EF&lt;/math&gt;. The hypotenuse of the triangle formed is the [[median]] of equilateral triangle &lt;math&gt;ADE&lt;/math&gt;, and one of the legs is &lt;math&gt;3\sqrt{2}&lt;/math&gt;. We apply the Pythagorean Theorem to deduce that the height is &lt;math&gt;6&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);<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 /> triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F); <br /> draw(B--Ba--Ca--C,dashed+d);<br /> draw(A--Aa--Da--D,dashed+d);<br /> draw(E--(E.x,E.y,0),dashed+l);<br /> draw(F--(F.x,F.y,0),dashed+l);<br /> draw(Aa--E--Da,dashed+d);<br /> draw(Ba--F--Ca,dashed+d);<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,NE);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> label(&quot;$12\sqrt{2}$&quot;,(E+F)/2,N);<br /> label(&quot;$6\sqrt{2}$&quot;,(A+B)/2,S);<br /> label(&quot;6&quot;,(3*s/2,s/2,3),ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Next, we complete the figure into a triangular prism, and find its volume, which is &lt;math&gt;\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432&lt;/math&gt;.<br /> <br /> Now, we subtract off the two extra [[pyramid]]s that we included, whose combined volume is &lt;math&gt;2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;432-144=\boxed{288}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); 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),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F);<br /> draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed);<br /> label(&quot;A&quot;,A,(-1,-1,0));<br /> label(&quot;B&quot;,B,( 2,-1,0));<br /> label(&quot;C&quot;,C,( 1, 1,0));<br /> label(&quot;D&quot;,D,(-1, 1,0));<br /> label(&quot;E&quot;,E,(0,0,1));<br /> label(&quot;F&quot;,F,(0,0,1));<br /> label(&quot;G&quot;,G,(0,0,-1));<br /> label(&quot;H&quot;,H,(0,0,-1));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Extend &lt;math&gt;EA&lt;/math&gt; and &lt;math&gt;FB&lt;/math&gt; to meet at &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; and &lt;math&gt;FC&lt;/math&gt; to meet at &lt;math&gt;H&lt;/math&gt;. Now, we have a regular tetrahedron &lt;math&gt;EFGH&lt;/math&gt;, which by symmetry has twice the volume of our original solid. This tetrahedron has side length &lt;math&gt;2s = 12\sqrt{2}&lt;/math&gt;. Using the formula for the volume of a regular tetrahedron, which is &lt;math&gt;V = \frac{\sqrt{2}S^3}{12}&lt;/math&gt;, where S is the side length of the tetrahedron, the volume of our original solid is:<br /> <br /> &lt;math&gt;V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is &lt;math&gt;6&lt;/math&gt;; thus, we will integrate with respect to height from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;, noting that each cross section of height &lt;math&gt;dh&lt;/math&gt; is a rectangle. The volume is then &lt;math&gt;\int_0^h(wl) \ \text{d}h&lt;/math&gt;, where &lt;math&gt;w&lt;/math&gt; is the width of the rectangle and &lt;math&gt;l&lt;/math&gt; is the length. We can express &lt;math&gt;w&lt;/math&gt; in terms of &lt;math&gt;h&lt;/math&gt; as &lt;math&gt;w=6\sqrt{2}-\sqrt{2}h&lt;/math&gt; since it decreases linearly with respect to &lt;math&gt;h&lt;/math&gt;, and &lt;math&gt;l=6\sqrt{2}+\sqrt{2}h&lt;/math&gt; since it similarly increases linearly with respect to &lt;math&gt;h&lt;/math&gt;. Now we solve:&lt;cmath&gt;\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}&lt;/cmath&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=Power_of_a_Point_Theorem&diff=104168 Power of a Point Theorem 2019-03-09T02:24:25Z <p>Zhang2018: /* Theorem */</p> <hr /> <div>The '''Power of a Point Theorem''' is a relationship that holds between the lengths of the [[line segment]]s formed when two [[line]]s [[intersect]] a [[circle]] and each other.<br /> <br /> == Theorem ==<br /> There are three possibilities as displayed in the figures below.<br /> <br /> # The two lines are [[secant line|chords]] of the circle and intersect inside the circle (figure on the left). In this case, we have &lt;math&gt; AE\cdot CE = BE\cdot DE &lt;/math&gt;.<br /> # One of the lines is [[tangent line|tangent]] to the circle while the other is a [[secant line|secant]] (middle figure). In this case, we have &lt;math&gt; AB^2 = BC\cdot BD &lt;/math&gt;.<br /> # Both lines are [[secant line|secants]] of the circle and intersect outside of it (figure on the right). In this case, we have &lt;math&gt; CB\cdot CA = CD\cdot CE. &lt;/math&gt;<br /> <br /> [[Image:Pop.PNG|center]]<br /> === Hint for Proof===<br /> Draw extra lines to create similar triangles! (Hint: Draw &lt;math&gt;AD&lt;/math&gt; on all three figures. Draw another line as well.)<br /> <br /> === Alternate Formulation ===<br /> This alternate formulation is much more compact, convenient, and general.<br /> <br /> Consider a circle &lt;math&gt;O&lt;/math&gt; and a point &lt;math&gt;P&lt;/math&gt; in the plane where &lt;math&gt;P&lt;/math&gt; is not on the circle. Now draw a line through &lt;math&gt;P&lt;/math&gt; that intersects the circle in two places. The power of a point theorem says that the product of the length from &lt;math&gt;P&lt;/math&gt; to the first point of intersection and the length from &lt;math&gt;P&lt;/math&gt; to the second point of intersection is constant for any choice of a line through &lt;math&gt;P&lt;/math&gt; that intersects the circle. This constant is called the power of point &lt;math&gt;P&lt;/math&gt;. For example, in the figure below <br /> &lt;cmath&gt;<br /> PX^2=PA_1\cdot PB_1=PA_2\cdot PB_2=\cdots=PA_i\cdot PB_i<br /> &lt;/cmath&gt;<br /> [[Image:Popalt.PNG|center]]<br /> <br /> Notice how this definition still works if &lt;math&gt;A_k&lt;/math&gt; and &lt;math&gt;B_k&lt;/math&gt; coincide (as is the case with &lt;math&gt;X&lt;/math&gt;). Consider also when &lt;math&gt;P&lt;/math&gt; is inside the circle. The definition still holds in this case.<br /> <br /> == Additional Notes ==<br /> One important result of this theorem is that both tangents from a point &lt;math&gt; P &lt;/math&gt; outside of a circle to that circle are equal in length.<br /> <br /> The theorem generalizes to higher dimensions, as follows.<br /> <br /> Let &lt;math&gt;P&lt;/math&gt; be a point, and let &lt;math&gt;S&lt;/math&gt; be an &lt;math&gt;n&lt;/math&gt;-sphere. Let two arbitrary lines passing through &lt;math&gt;P&lt;/math&gt; intersect &lt;math&gt;S&lt;/math&gt; at &lt;math&gt;A_1,B_1;A_2,B_2&lt;/math&gt;, respectively. Then<br /> &lt;cmath&gt;<br /> PA_1\cdot PB_1=PA_2\cdot PB_2<br /> &lt;/cmath&gt;<br /> <br /> ''Proof.'' We have already proven the theorem for a &lt;math&gt;1&lt;/math&gt;-sphere (i.e., a circle), so it only remains to prove the theorem for more dimensions. Consider the [[plane]] &lt;math&gt;p&lt;/math&gt; containing both of the lines passing through &lt;math&gt;P&lt;/math&gt;. The intersection of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;S&lt;/math&gt; must be a circle. If we consider the lines and &lt;math&gt;P&lt;/math&gt; with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.<br /> <br /> == Problems ==<br /> The problems are divided into three categories: introductory, intermediate, and olympiad.<br /> <br /> === Introductory ===<br /> ==== Problem 1 ====<br /> Find the value of &lt;math&gt;x&lt;/math&gt; in the following diagram:<br /> <br /> [[Image:popprob1.PNG|center]]<br /> <br /> [[Power of a Point Theorem/Introductory_Problem_1|Solution]]<br /> <br /> ==== Problem 2 ==== <br /> Find the value of &lt;math&gt;x&lt;/math&gt; in the following diagram:<br /> <br /> [[Image:popprob2.PNG|center]]<br /> <br /> [[Power of a Point Theorem/Introductory_Problem_2|Solution]]<br /> <br /> ==== Problem 3 ==== <br /> ([[ARML]]) In a circle, chords &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; intersect at &lt;math&gt;R&lt;/math&gt;. If &lt;math&gt;AR:BR=1:4&lt;/math&gt; and &lt;math&gt;CR:DR=4:9&lt;/math&gt;, find the ratio &lt;math&gt;AB:CD&lt;/math&gt;.<br /> <br /> [[Image:popprob3.PNG|center]]<br /> <br /> [[Power of a Point Theorem/Introductory_Problem_3|Solution]]<br /> <br /> ==== Problem 4 ====<br /> ([[ARML]]) Chords &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; of a given circle are [[perpendicular]] to each other and intersect at a right angle at point &lt;math&gt;E&lt;/math&gt;. Given that &lt;math&gt;BE=16&lt;/math&gt;, &lt;math&gt;DE=4&lt;/math&gt;, and &lt;math&gt;AD=5&lt;/math&gt;, find &lt;math&gt;CE&lt;/math&gt;.<br /> <br /> [[Power of a Point Theorem/Introductory_Problem_4|Solution]]<br /> <br /> === Intermediate ===<br /> ==== Problem 1 ====<br /> Two tangents from an external point &lt;math&gt;P&lt;/math&gt; are drawn to a circle and intersect it at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third tangent meets the circle at &lt;math&gt;T&lt;/math&gt;, and the tangents &lt;math&gt;\overrightarrow{PA}&lt;/math&gt; and &lt;math&gt;\overrightarrow{PB}&lt;/math&gt; at points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, respectively. Find the perimeter of &lt;math&gt;\triangle PQR&lt;/math&gt;.<br /> <br /> ==== Problem 2 ====<br /> Square &lt;math&gt;ABCD&lt;/math&gt; of side length &lt;math&gt;10&lt;/math&gt; has a circle inscribed in it. Let &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AB}.&lt;/math&gt; Find the length of that portion of the segment &lt;math&gt;\overline{MC}&lt;/math&gt; that lies outside of the circle.<br /> <br /> ==== Other Intermediate Example Problems ====<br /> * [[1971_Canadian_MO_Problems/Problem_1 | 1971 Canadian Mathematics Olympiad Problem 1]]<br /> * [[2005 AIME I Problems/Problem 15|2005 AIME I Problem 15]]<br /> <br /> ==See also==<br /> * [[Geometry]]<br /> * [[Planar figures]]<br /> <br /> [[Category:Geometry]]<br /> [[Category:Theorems]]</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems&diff=100802 2018 AMC 8 Problems 2019-01-24T01:29:35Z <p>Zhang2018: /* Problem 12 */</p> <hr /> <div>==Problem 1==<br /> An amusement park has a collection of scale models, with ratio &lt;math&gt;1 : 20&lt;/math&gt;, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica to the nearest whole number?<br /> <br /> &lt;math&gt;\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> What is the value of the product&lt;cmath&gt;\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 2|Solution]]<br /> ==Problem 3==<br /> Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?<br /> <br /> &lt;math&gt;\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 3|Solution]]<br /> ==Problem 4==<br /> The twelve-sided figure shown has been drawn on &lt;math&gt;1 \text{ cm}\times 1 \text{ cm}&lt;/math&gt; graph paper. What is the area of the figure in &lt;math&gt;\text{cm}^2&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(8mm);<br /> for (int i=0; i&lt;7; ++i) {<br /> draw((i,0)--(i,7),gray);<br /> draw((0,i+1)--(7,i+1),gray);<br /> }<br /> draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 4|Solution]]<br /> ==Problem 5==<br /> What is the value of &lt;math&gt;1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 5|Solution]]<br /> ==Problem 6==<br /> On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?<br /> <br /> &lt;math&gt;\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 6|Solution]]<br /> ==Problem 7==<br /> The &lt;math&gt;5&lt;/math&gt;-digit number &lt;math&gt;\underline{2}&lt;/math&gt; &lt;math&gt;\underline{0}&lt;/math&gt; &lt;math&gt;\underline{1}&lt;/math&gt; &lt;math&gt;\underline{8}&lt;/math&gt; &lt;math&gt;\underline{U}&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. What is the remainder when this number is divided by &lt;math&gt;8&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> void drawbar(real x, real h) {<br /> fill((x-0.15,0.5)--(x+0.15,0.5)--(x+0.15,h)--(x-0.15,h)--cycle,gray);<br /> }<br /> draw((0.5,0.5)--(7.5,0.5)--(7.5,5)--(0.5,5)--cycle);<br /> for (real i=1; i&lt;5; i=i+0.5) {<br /> draw((0.5,i)--(7.5,i),gray);<br /> }<br /> drawbar(1.0,1.0);<br /> drawbar(2.0,2.0);<br /> drawbar(3.0,1.5);<br /> drawbar(4.0,3.5);<br /> drawbar(5.0,4.5);<br /> drawbar(6.0,2.0);<br /> drawbar(7.0,1.5);<br /> for (int i=1; i&lt;8; ++i) {<br /> label(&quot;$&quot;+string(i)+&quot;$&quot;,(i,0.25));<br /> }<br /> for (int i=1; i&lt;9; ++i) {<br /> label(&quot;$&quot;+string(i)+&quot;$&quot;,(0.5,0.5*(i+1)),W);<br /> }<br /> label(&quot;Number of Days of Exercise&quot;,(4,-0.1));<br /> label(rotate(90)*&quot;Number of Students&quot;,(-0.1,2.75));<br /> &lt;/asy&gt;<br /> What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?<br /> <br /> &lt;math&gt;\textbf{(A) } 3.50 \qquad \textbf{(B) } 3.57 \qquad \textbf{(C) } 4.36 \qquad \textbf{(D) } 4.50 \qquad \textbf{(E) } 5.00&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 8|Solution]]<br /> ==Problem 9==<br /> Tyler is tiling the floor of his 12 foot by 16 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }91\qquad\textbf{(D) }96\qquad \textbf{(E) }120&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> The &lt;math&gt;\emph{harmonic mean}&lt;/math&gt; of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{3}{7}\qquad\textbf{(B) }\frac{7}{12}\qquad\textbf{(C) }\frac{12}{7}\qquad\textbf{(D) }\frac{7}{4}\qquad \textbf{(E) }\frac{7}{3}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \text{X}&amp;\quad\text{X}\quad&amp;\text{X} \\<br /> \text{X}&amp;\quad\text{X}\quad&amp;\text{X} <br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual time?<br /> <br /> &lt;math&gt;\textbf{(A) }5:50\qquad\textbf{(B) }6:00\qquad\textbf{(C) }6:30\qquad\textbf{(D) }6:55\qquad \textbf{(E) }8:10&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Let &lt;math&gt;N&lt;/math&gt; be the greatest five-digit number whose digits have a product of &lt;math&gt;120&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 14|Solution]]<br /> ==Problem 15==<br /> In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of &lt;math&gt;1&lt;/math&gt; square unit, then what is the area of the shaded region, in square units?<br /> <br /> &lt;asy&gt;<br /> size(4cm);<br /> filldraw(scale(2)*unitcircle,gray,black);<br /> filldraw(shift(-1,0)*unitcircle,white,black);<br /> filldraw(shift(1,0)*unitcircle,white,black);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 15|Solution]]<br /> ==Problem 16==<br /> Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?<br /> <br /> &lt;math&gt;\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 16|Solution]]<br /> ==Problem 17==<br /> Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is &lt;math&gt;2&lt;/math&gt; miles, which is &lt;math&gt;10,560&lt;/math&gt; feet, and Bella covers &lt;math&gt;2 \tfrac{1}{2}&lt;/math&gt; feet with each step. How many steps will Bella take by the time she meets Ella?<br /> <br /> &lt;math&gt;\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 17|Solution]]<br /> ==Problem 18==<br /> How many positive factors does &lt;math&gt;23,232&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 18|Solution]]<br /> ==Problem 19==<br /> In a sign pyramid a cell gets a &quot;+&quot; if the two cells below it have the same sign, and it gets a &quot;-&quot; if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a &quot;+&quot; at the top of the pyramid?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle;<br /> draw(box); label(&quot;$+$&quot;,(0,0));<br /> draw(shift(1,0)*box); label(&quot;$-$&quot;,(1,0));<br /> draw(shift(2,0)*box); label(&quot;$+$&quot;,(2,0));<br /> draw(shift(3,0)*box); label(&quot;$-$&quot;,(3,0));<br /> draw(shift(0.5,0.4)*box); label(&quot;$-$&quot;,(0.5,0.4));<br /> draw(shift(1.5,0.4)*box); label(&quot;$-$&quot;,(1.5,0.4));<br /> draw(shift(2.5,0.4)*box); label(&quot;$-$&quot;,(2.5,0.4));<br /> draw(shift(1,0.8)*box); label(&quot;$+$&quot;,(1,0.8));<br /> draw(shift(2,0.8)*box); label(&quot;$+$&quot;,(2,0.8));<br /> draw(shift(1.5,1.2)*box); label(&quot;$+$&quot;,(1.5,1.2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 19|Solution]]<br /> ==Problem 20==<br /> In &lt;math&gt;\triangle ABC,&lt;/math&gt; a point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;AE=1&lt;/math&gt; and &lt;math&gt;EB=2.&lt;/math&gt; Point &lt;math&gt;D&lt;/math&gt; is on &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;\overline{DE} \parallel \overline{BC}&lt;/math&gt; and point &lt;math&gt;F&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;\overline{EF} \parallel \overline{AC}.&lt;/math&gt; What is the ratio of the area of &lt;math&gt;CDEF&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(7cm);<br /> pair A,B,C,DD,EE,FF;<br /> A = (0,0); B = (3,0); C = (0.5,2.5);<br /> EE = (1,0);<br /> DD = intersectionpoint(A--C,EE--EE+(C-B));<br /> FF = intersectionpoint(B--C,EE--EE+(C-A));<br /> draw(A--B--C--A--DD--EE--FF,black+1bp);<br /> label(&quot;$A$&quot;,A,S); label(&quot;$B$&quot;,B,S); label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,DD,W); label(&quot;$E$&quot;,EE,S); label(&quot;$F$&quot;,FF,NE);<br /> label(&quot;$1$&quot;,(A+EE)/2,S); label(&quot;$2$&quot;,(EE+B)/2,S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 20|Solution]]<br /> ==Problem 21==<br /> How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 21|Solution]]<br /> ==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?<br /> <br /> &lt;asy&gt;<br /> size(3cm);<br /> pair A[];<br /> for (int i=0; i&lt;9; ++i) {<br /> A[i] = rotate(22.5+45*i)*(1,0);<br /> }<br /> filldraw(A--A--A--A--A--A--A--A--cycle,gray,black);<br /> for (int i=0; i&lt;8; ++i) { dot(A[i]); }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> In the cube &lt;math&gt;ABCDEFGH&lt;/math&gt; with opposite vertices &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E,&lt;/math&gt; &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt; are the midpoints of edges &lt;math&gt;\overline{FB}&lt;/math&gt; and &lt;math&gt;\overline{HD},&lt;/math&gt; respectively. Let &lt;math&gt;R&lt;/math&gt; be the ratio of the area of the cross-section &lt;math&gt;EJCI&lt;/math&gt; to the area of one of the faces of the cube. What is &lt;math&gt;R^2?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> pair A,B,C,D,EE,F,G,H,I,J;<br /> C = (0,0);<br /> B = (-1,1);<br /> D = (2,0.5);<br /> A = B+D;<br /> G = (0,2);<br /> F = B+G;<br /> H = G+D;<br /> EE = G+B+D;<br /> I = (D+H)/2; J = (B+F)/2;<br /> filldraw(C--I--EE--J--cycle,lightgray,black);<br /> draw(C--D--H--EE--F--B--cycle); <br /> draw(G--F--G--C--G--H);<br /> draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed);<br /> dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J);<br /> label(&quot;$A$&quot;,A,E);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,S);<br /> label(&quot;$D$&quot;,D,E);<br /> label(&quot;$E$&quot;,EE,N);<br /> label(&quot;$F$&quot;,F,W);<br /> label(&quot;$G$&quot;,G,N);<br /> label(&quot;$H$&quot;,H,E);<br /> label(&quot;$I$&quot;,I,E);<br /> label(&quot;$J$&quot;,J,W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> How many perfect cubes lie between &lt;math&gt;2^8+1&lt;/math&gt; and &lt;math&gt;2^{18}+1&lt;/math&gt;, inclusive?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> {{MAA Notice}}<br /> <br /> <br /> Link back to original page: https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8</div> Zhang2018 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems&diff=100801 2018 AMC 8 Problems 2019-01-24T01:28:10Z <p>Zhang2018: /* Problem 12 */</p> <hr /> <div>==Problem 1==<br /> An amusement park has a collection of scale models, with ratio &lt;math&gt;1 : 20&lt;/math&gt;, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica to the nearest whole number?<br /> <br /> &lt;math&gt;\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> What is the value of the product&lt;cmath&gt;\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 2|Solution]]<br /> ==Problem 3==<br /> Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?<br /> <br /> &lt;math&gt;\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 3|Solution]]<br /> ==Problem 4==<br /> The twelve-sided figure shown has been drawn on &lt;math&gt;1 \text{ cm}\times 1 \text{ cm}&lt;/math&gt; graph paper. What is the area of the figure in &lt;math&gt;\text{cm}^2&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(8mm);<br /> for (int i=0; i&lt;7; ++i) {<br /> draw((i,0)--(i,7),gray);<br /> draw((0,i+1)--(7,i+1),gray);<br /> }<br /> draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 4|Solution]]<br /> ==Problem 5==<br /> What is the value of &lt;math&gt;1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 5|Solution]]<br /> ==Problem 6==<br /> On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?<br /> <br /> &lt;math&gt;\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 6|Solution]]<br /> ==Problem 7==<br /> The &lt;math&gt;5&lt;/math&gt;-digit number &lt;math&gt;\underline{2}&lt;/math&gt; &lt;math&gt;\underline{0}&lt;/math&gt; &lt;math&gt;\underline{1}&lt;/math&gt; &lt;math&gt;\underline{8}&lt;/math&gt; &lt;math&gt;\underline{U}&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. What is the remainder when this number is divided by &lt;math&gt;8&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> void drawbar(real x, real h) {<br /> fill((x-0.15,0.5)--(x+0.15,0.5)--(x+0.15,h)--(x-0.15,h)--cycle,gray);<br /> }<br /> draw((0.5,0.5)--(7.5,0.5)--(7.5,5)--(0.5,5)--cycle);<br /> for (real i=1; i&lt;5; i=i+0.5) {<br /> draw((0.5,i)--(7.5,i),gray);<br /> }<br /> drawbar(1.0,1.0);<br /> drawbar(2.0,2.0);<br /> drawbar(3.0,1.5);<br /> drawbar(4.0,3.5);<br /> drawbar(5.0,4.5);<br /> drawbar(6.0,2.0);<br /> drawbar(7.0,1.5);<br /> for (int i=1; i&lt;8; ++i) {<br /> label(&quot;$&quot;+string(i)+&quot;$&quot;,(i,0.25));<br /> }<br /> for (int i=1; i&lt;9; ++i) {<br /> label(&quot;$&quot;+string(i)+&quot;$&quot;,(0.5,0.5*(i+1)),W);<br /> }<br /> label(&quot;Number of Days of Exercise&quot;,(4,-0.1));<br /> label(rotate(90)*&quot;Number of Students&quot;,(-0.1,2.75));<br /> &lt;/asy&gt;<br /> What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?<br /> <br /> &lt;math&gt;\textbf{(A) } 3.50 \qquad \textbf{(B) } 3.57 \qquad \textbf{(C) } 4.36 \qquad \textbf{(D) } 4.50 \qquad \textbf{(E) } 5.00&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 8|Solution]]<br /> ==Problem 9==<br /> Tyler is tiling the floor of his 12 foot by 16 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }91\qquad\textbf{(D) }96\qquad \textbf{(E) }120&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> The &lt;math&gt;\emph{harmonic mean}&lt;/math&gt; of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{3}{7}\qquad\textbf{(B) }\frac{7}{12}\qquad\textbf{(C) }\frac{12}{7}\qquad\textbf{(D) }\frac{7}{4}\qquad \textbf{(E) }\frac{7}{3}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \text{X}&amp;\quad\text{X}\quad&amp;\text{X} \\<br /> \text{X}&amp;\quad\text{X}\quad&amp;\text{X} <br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual tim?<br /> <br /> &lt;math&gt;\textbf{(A) }5:50\qquad\textbf{(B) }6:00\qquad\textbf{(C) }6:30\qquad\textbf{(D) }6:55\qquad \textbf{(E) }8:10&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Let &lt;math&gt;N&lt;/math&gt; be the greatest five-digit number whose digits have a product of &lt;math&gt;120&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 14|Solution]]<br /> ==Problem 15==<br /> In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of &lt;math&gt;1&lt;/math&gt; square unit, then what is the area of the shaded region, in square units?<br /> <br /> &lt;asy&gt;<br /> size(4cm);<br /> filldraw(scale(2)*unitcircle,gray,black);<br /> filldraw(shift(-1,0)*unitcircle,white,black);<br /> filldraw(shift(1,0)*unitcircle,white,black);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 15|Solution]]<br /> ==Problem 16==<br /> Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?<br /> <br /> &lt;math&gt;\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 16|Solution]]<br /> ==Problem 17==<br /> Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is &lt;math&gt;2&lt;/math&gt; miles, which is &lt;math&gt;10,560&lt;/math&gt; feet, and Bella covers &lt;math&gt;2 \tfrac{1}{2}&lt;/math&gt; feet with each step. How many steps will Bella take by the time she meets Ella?<br /> <br /> &lt;math&gt;\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 17|Solution]]<br /> ==Problem 18==<br /> How many positive factors does &lt;math&gt;23,232&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 18|Solution]]<br /> ==Problem 19==<br /> In a sign pyramid a cell gets a &quot;+&quot; if the two cells below it have the same sign, and it gets a &quot;-&quot; if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a &quot;+&quot; at the top of the pyramid?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle;<br /> draw(box); label(&quot;$+$&quot;,(0,0));<br /> draw(shift(1,0)*box); label(&quot;$-$&quot;,(1,0));<br /> draw(shift(2,0)*box); label(&quot;$+$&quot;,(2,0));<br /> draw(shift(3,0)*box); label(&quot;$-$&quot;,(3,0));<br /> draw(shift(0.5,0.4)*box); label(&quot;$-$&quot;,(0.5,0.4));<br /> draw(shift(1.5,0.4)*box); label(&quot;$-$&quot;,(1.5,0.4));<br /> draw(shift(2.5,0.4)*box); label(&quot;$-$&quot;,(2.5,0.4));<br /> draw(shift(1,0.8)*box); label(&quot;$+$&quot;,(1,0.8));<br /> draw(shift(2,0.8)*box); label(&quot;$+$&quot;,(2,0.8));<br /> draw(shift(1.5,1.2)*box); label(&quot;$+$&quot;,(1.5,1.2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 19|Solution]]<br /> ==Problem 20==<br /> In &lt;math&gt;\triangle ABC,&lt;/math&gt; a point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;AE=1&lt;/math&gt; and &lt;math&gt;EB=2.&lt;/math&gt; Point &lt;math&gt;D&lt;/math&gt; is on &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;\overline{DE} \parallel \overline{BC}&lt;/math&gt; and point &lt;math&gt;F&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;\overline{EF} \parallel \overline{AC}.&lt;/math&gt; What is the ratio of the area of &lt;math&gt;CDEF&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(7cm);<br /> pair A,B,C,DD,EE,FF;<br /> A = (0,0); B = (3,0); C = (0.5,2.5);<br /> EE = (1,0);<br /> DD = intersectionpoint(A--C,EE--EE+(C-B));<br /> FF = intersectionpoint(B--C,EE--EE+(C-A));<br /> draw(A--B--C--A--DD--EE--FF,black+1bp);<br /> label(&quot;$A$&quot;,A,S); label(&quot;$B$&quot;,B,S); label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,DD,W); label(&quot;$E$&quot;,EE,S); label(&quot;$F$&quot;,FF,NE);<br /> label(&quot;$1$&quot;,(A+EE)/2,S); label(&quot;$2$&quot;,(EE+B)/2,S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 20|Solution]]<br /> ==Problem 21==<br /> How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 21|Solution]]<br /> ==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?<br /> <br /> &lt;asy&gt;<br /> size(3cm);<br /> pair A[];<br /> for (int i=0; i&lt;9; ++i) {<br /> A[i] = rotate(22.5+45*i)*(1,0);<br /> }<br /> filldraw(A--A--A--A--A--A--A--A--cycle,gray,black);<br /> for (int i=0; i&lt;8; ++i) { dot(A[i]); }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> In the cube &lt;math&gt;ABCDEFGH&lt;/math&gt; with opposite vertices &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E,&lt;/math&gt; &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt; are the midpoints of edges &lt;math&gt;\overline{FB}&lt;/math&gt; and &lt;math&gt;\overline{HD},&lt;/math&gt; respectively. Let &lt;math&gt;R&lt;/math&gt; be the ratio of the area of the cross-section &lt;math&gt;EJCI&lt;/math&gt; to the area of one of the faces of the cube. What is &lt;math&gt;R^2?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> pair A,B,C,D,EE,F,G,H,I,J;<br /> C = (0,0);<br /> B = (-1,1);<br /> D = (2,0.5);<br /> A = B+D;<br /> G = (0,2);<br /> F = B+G;<br /> H = G+D;<br /> EE = G+B+D;<br /> I = (D+H)/2; J = (B+F)/2;<br /> filldraw(C--I--EE--J--cycle,lightgray,black);<br /> draw(C--D--H--EE--F--B--cycle); <br /> draw(G--F--G--C--G--H);<br /> draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed);<br /> dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J);<br /> label(&quot;$A$&quot;,A,E);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,S);<br /> label(&quot;$D$&quot;,D,E);<br /> label(&quot;$E$&quot;,EE,N);<br /> label(&quot;$F$&quot;,F,W);<br /> label(&quot;$G$&quot;,G,N);<br /> label(&quot;$H$&quot;,H,E);<br /> label(&quot;$I$&quot;,I,E);<br /> label(&quot;$J$&quot;,J,W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> How many perfect cubes lie between &lt;math&gt;2^8+1&lt;/math&gt; and &lt;math&gt;2^{18}+1&lt;/math&gt;, inclusive?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58&lt;/math&gt;<br /> <br /> [[2018 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> {{MAA Notice}}<br /> <br /> <br /> Link back to original page: https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8</div> Zhang2018