https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Dipenm&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T08:12:05ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_17&diff=903312017 AMC 10B Problems/Problem 172018-02-07T01:03:41Z<p>Dipenm: /* See Also */</p>
<hr />
<div>{{duplicate|[[2017 AMC 12B Problems|2017 AMC 12B #11]] and [[2017 AMC 10B Problems|2017 AMC 10B #17]]}}<br />
<br />
==Problem==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
==Solution 1==<br />
<br />
Case 1: monotonous numbers with digits in ascending order<br />
<br />
There are <math>\Sigma_{n=1}^{9} \binom{9}{n}</math> ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, <math>\emptyset</math> (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to <math>\Sigma_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.</math><br />
<br />
Case 2: monotonous numbers with digits in descending order<br />
<br />
There are <math>\Sigma_{n=1}^{10} \binom{10}{n}</math> ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. However, <math>\emptyset</math> (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to <math>\Sigma_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.</math> We discard the number 0 since it is not positive. Thus there are <math>1022</math> here. <br />
<br />
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are <math>511+1022-9=\boxed{\textbf{(B)} 1524}</math> monotonous numbers.<br />
<br />
==Solution 2==<br />
Like Solution 1, divide the problem into an increasing and decreasing case:<br />
<br />
<math>\bullet</math> Case 1: Monotonous numbers with digits in ascending order.<br />
<br />
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.<br />
<br />
To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are <math>2^9 = 512</math> ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get <math>512-1=511</math> monotonous numbers for this case.<br />
<br />
<math>\bullet</math> Case 2: Monotonous numbers with digits in descending order.<br />
<br />
This time, we arrange all 10 digits in decreasing order and repeat the process to find <math>2^{10} = 1024</math> ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get <math>1024-2=1022</math> monotonous numbers for this case.<br />
<br />
At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.<br />
<br />
Thus our final answer is <math>511+1022-9 = \boxed{\textbf{(B) } 1524}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=16|num-a=18}}<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=12}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_7&diff=903302017 AMC 10B Problems/Problem 72018-02-07T01:02:23Z<p>Dipenm: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
==Solution 1==<br />
Let's call the distance that Samia had to travel in total as <math>2x</math>, so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both <math>\frac{2x}{2}</math>, or <math>x</math>.<br />
<cmath></cmath><br />
She bikes at a rate of <math>17</math> kph, so she travels the distance she bikes in <math>\frac{x}{17}</math> hours. She walks at a rate of <math>5</math> kph, so she travels the distance she walks in <math>\frac{x}{5}</math> hours.<br />
<cmath></cmath><br />
The total time is <math>\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}</math>. This is equal to <math>\frac{44}{60} = \frac{11}{15}</math> of an hour. Solving for <math>x</math>, we have:<br />
<cmath></cmath><br />
<cmath>\frac{22x}{85} = \frac{11}{15}</cmath><br />
<cmath>\frac{2x}{85} = \frac{1}{15}</cmath><br />
<cmath>30x = 85</cmath><br />
<cmath>6x = 17</cmath><br />
<cmath>x = \frac{17}{6}</cmath><br />
<cmath></cmath><br />
Since <math>x</math> is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about <math>\boxed{\bold{(C)} 2.8}</math>.<br />
{{AMC10 box|year=2017|ab=B|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=5}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_4&diff=903272017 AMC 10B Problems/Problem 42018-02-07T01:01:31Z<p>Dipenm: /* Solutions */</p>
<hr />
<div>==Problem==<br />
Supposed that <math>x</math> and <math>y</math> are nonzero real numbers such that <math>\frac{3x+y}{x-3y}=-2</math>. What is the value of <math>\frac{x+3y}{3x-y}</math>?<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3</math><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
Rearranging, we find <math>3x+y=-2x+6y</math>, or <math>5x=5y\implies x=y</math>.<br />
Substituting, we can convert the second equation into <math>\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}</math>.<br />
<br />
===Solution 2===<br />
Substituting each <math>x</math> and <math>y</math> with <math>1</math>, we see that the given equation holds true, as <math>\frac{3(1)+1}{1-3(1)} = -2</math>. Thus, <math>\frac{x+3y}{3x-y}=\boxed{\textbf{(D)}\ 2}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}}<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=4}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_3&diff=903242017 AMC 10B Problems/Problem 32018-02-07T00:59:44Z<p>Dipenm: /* See Also */</p>
<hr />
<div>==Problem==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satisfy the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
==Solution==<br />
Notice that <math>y+z</math> must be positive because <math>|z|>|y|</math>. Therefore the answer is <math>\boxed{\textbf{(E) } y+z}</math>.<br />
<br />
The other choices:<br />
<br />
<math>\textbf{(A)}</math> As <math>x</math> grows closer to <math>0</math>, <math>x^2</math> decreases and thus becomes less than <math>y</math>.<br />
<br />
<math>\textbf{(B)}</math> <math>x</math> can be as small as possible (<math>x>0</math>), so <math>xz</math> grows close to <math>0</math> as <math>x</math> approaches <math>0</math>.<br />
<br />
<math>\textbf{(C)}</math> For all <math>-1<y<0</math>, <math>y>y^2</math>, and thus it is always negative.<br />
<br />
<math>\textbf{(D)}</math> The same logic as above, but when <math>-\frac{1}{2}<y<0</math> this time.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=3}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_3&diff=846592010 AIME I Problems/Problem 32017-03-12T01:17:58Z<p>Dipenm: </p>
<hr />
<div>== Problem ==<br />
Suppose that <math>y = \frac34x</math> and <math>x^y = y^x</math>. The quantity <math>x + y</math> can be expressed as a rational number <math>\frac {r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r + s</math>.<br />
<br />
== Solution ==<br />
We solve in general using <math>c</math> instead of <math>3/4</math>. Substituting <math>y = cx</math>, we have:<br />
<br />
<center><cmath>x^{cx} = (cx)^x \Longrightarrow (x^x)^c = c^x\cdot x^x</cmath></center><br />
<br />
Dividing by <math>x^x</math>, we get <math>(x^x)^{c - 1} = c^x</math>.<br />
<br />
Taking the <math>x</math>th root, <math>x^{c - 1} = c</math>, or <math>x = c^{1/(c - 1)}</math>.<br />
<br />
In the case <math>c = \frac34</math>, <math>x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}</math>, <math>y = \frac {64}{27}</math>, <math>x + y = \frac {256 + 192}{81} = \frac {448}{81}</math>, yielding an answer of <math>448 + 81 = \boxed{529}</math>.<br />
<br />
== Solution 2 ==<br />
Taking the logarithm base <math>x</math> of both sides, we arrive with:<br />
<br />
<center><cmath> y = log_x y^x \Longrightarrow \frac{y}{x} = log_x y = log_x \frac{3}{4}x = \frac{3}{4}</cmath></center><br />
Where the last two simplifications were made since <math>y = \frac{3}{4}x</math>. Then,<br />
<center><cmath>x^{\frac{3}{4}} = \frac{3}{4}x \Longrightarrow x^{\frac{1}{4}} = \frac{4}{3} \Longrightarrow x = \left(\frac{4}{3}\right)^4</cmath></center><br />
Then, <math>y = \left(\frac{4}{3}\right)^3</math>, and thus:<br />
<center> <cmath>x+y = \left(\frac{4}{3}\right)^3 \left(\frac{4}{3} + 1 \right) = \frac{448}{81} \Longrightarrow 448 + 81 = \boxed{529}</cmath> </center><br />
<br />
== See Also ==<br />
{{AIME box|year=2010|num-b=2|num-a=4|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_7&diff=826292016 AMC 12A Problems/Problem 72017-01-30T03:40:08Z<p>Dipenm: /* Problem */</p>
<hr />
<div>==Problem==<br />
Which of these describes the graph of <math>x^2(x+y+1)=y^2(x+y+1)</math> ?<br />
<br />
<math>\textbf{(A)}\ \text{two parallel lines}\\ \qquad\textbf{(B)}\ \text{two intersecting lines}\\ \qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\\ \qquad\textbf{(D)}\ \text{three lines that do not all pass through a common point}\\ \qquad\textbf{(E)}\ \text{a line and a parabola}</math><br />
<br />
==Solution 1==<br />
<br />
The equation <math>x^2(x+y+1)=y^2(x+y+1)</math> tells us <math>x^2=y^2</math> or <math>x+y+1=0</math> . <math>x^2=y^2</math> generates two lines <math>y=x</math> and <math>y=-x</math> . <math>x+y+1=0</math> is another straight line. The only intersection of <math>y=x</math> and <math>y=-x</math> is <math>(0,0)</math> , which is not on <math>x+y+1=0</math> . Therefore, the graph is three lines that do not have a common intersection, or <math>\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}</math><br />
<br />
==Solution 2==<br />
<br />
If <math>x+y+1\neq0</math>, then dividing both sides of the equation by <math>x+y+1</math> gives us <math>x^2=y^2</math>. Rearranging and factoring, we get <math>x^2-y^2=(x+y)(x-y)=0</math>. If <math>x+y+1=0</math>, then the equation is satisfied. Thus either <math>x+y=0</math>, <math>x-y=0</math>, or <math>x+y+1=0</math>. These equations can be rearranged into the lines <math>y=-x</math>, <math>y=x</math>, and <math>y=-x-1</math>, respectively. Since these three lines are distinct, the answer is <math>\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}</math>.<br />
<br />
==Solution 3==<br />
<br />
Subtract <math>y^2(x+y+1)</math> on both sides of the equation to get <math>x^2(x+y+1)-y^2(x+y+1)=0</math>. Factoring <math>x+y+1</math> gives us <math>(x+y+1)(x^2-y^2)=(x+y+1)(x+y)(x-y)=0</math>, so either <math>x+y+1=0</math>, <math>x+y=0</math>, or <math>x-y=0</math>. Continue on with the second half of solution 2.<br />
<br />
==Diagram:==<br />
<br />
<math>AB: y=x</math><br />
<br />
<math>CD: y=-x</math><br />
<br />
<math>EF: x+y+1=0</math><br />
<br />
<asy><br />
size(7cm);<br />
pair F= (5,0), E=(-1,6), D=(0,0), C=(6,0), B=(6,6), A=(0,6);<br />
draw(A--C);<br />
draw(B--D);<br />
draw(E--F);<br />
<br />
label("$A$", A, dir(135));<br />
label("$B$", C, dir(-45));<br />
label("$C$", B, dir(45));<br />
label("$D$", D, dir(-135));<br />
label("$E$", E, dir(135));<br />
label("$F$", F, dir(-45));<br />
</asy><br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=A|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_20&diff=763372011 AMC 10B Problems/Problem 202016-02-16T22:07:21Z<p>Dipenm: /* Solution */</p>
<hr />
<div>{{duplicate|[[2011 AMC 12B Problems|2011 AMC 12B #16]] and [[2011 AMC 10B Problems|2011 AMC 10B #20]]}}<br />
<br />
== Problem==<br />
<br />
Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120</math>°. Region <math>R</math> consists of all points inside the rhombus that are closer to vertex <math>B</math> than any of the other three vertices. What is the area of <math>R</math>?<br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
== Solution ==<br />
<br />
Suppose that <math>P</math> is a point in the rhombus <math>ABCD</math> and let <math>\ell_{BC}</math> be the [[perpendicular bisector]] of <math>\overline{BC}</math>. Then <math>PB < PC</math> if and only if <math>P</math> is on the same side of <math>\ell_{BC}</math> as <math>B</math>. The line <math>\ell_{BC}</math> divides the plane into two half-planes; let <math>S_{BC}</math> be the half-plane containing <math>B</math>. Let us define similarly <math>\ell_{BD},S_{BD}</math> and <math>\ell_{BA},S_{BA}</math>. Then <math>R</math> is equal to <math>ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}</math>. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:<br />
<br />
<asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(0.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2;<br />
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray);<br />
draw(A--B--C--D--cycle);<br />
draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G);<br />
<br />
label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW);<br />
label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE);<br />
label("$2$",(D--C),SW);<br />
</asy><br />
Since <math>\triangle BCD</math> and <math>\triangle BAD</math> are equilateral, <math>\ell_{BC}</math> contains <math>D</math>, <math>\ell_{BD}</math> contains <math>A</math> and <math>C</math>, and <math>\ell_{BA}</math> contains <math>D</math>. Then <math>\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH</math> with <math>BE = 1</math> and <math>EF = \frac{1}{\sqrt{3}}</math> so <math>[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}</math>. Multiply this by 4 and it turns out that the pentagon has area <math>\boxed{(C)\frac{2\sqrt{3}}{3}}</math>.<br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=19|num-a=21}}<br />
<br />
{{AMC12 box|year=2011|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_15&diff=749642011 AMC 10A Problems/Problem 152016-01-31T17:17:18Z<p>Dipenm: </p>
<hr />
<div>== Problem 15 ==<br />
<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
'''(A) 140 (B) 240 (C) 440 (D) 640 (E) 840'''<br />
<br />
== Solution ==<br />
<br />
We know that <math>\frac{\text{total miles}}{\text{total gas}}=55</math>. Let <math>x</math> be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is <math>0.02x</math>. The total distance traveled is <math>40+x</math>, so we get <math>\frac{40+x}{0.02x}=55</math>. Solving this equation, we get <math>x=400</math>, so the total distance is <math>400 + 40 = \boxed{440 \ \mathbf{(C)}}</math>.<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2011|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_1&diff=749632012 AMC 10A Problems/Problem 12016-01-31T16:27:52Z<p>Dipenm: /* Solution */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #2]] and [[2012 AMC 10A Problems|2012 AMC 10A #1]]}}<br />
<br />
== Problem ==<br />
<br />
Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?<br />
<br />
<math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30 </math><br />
<br />
== Solution 1 ==<br />
<br />
Cagney can frost one in <math>20</math> seconds, and Lacey can frost one in <math>30</math> seconds. Working together, they can frost one in <math>\frac{20\cdot30}{20+30} = \frac{600}{50} = 12</math> seconds. In <math>300</math> seconds (<math>5</math> minutes), they can frost <math>\boxed{\textbf{(D)}\ 25}</math> cupcakes.<br />
<br />
== Solution 2 ==<br />
<br />
In <math>300</math> seconds (<math>5</math> minutes), Cagney will frost <math>\dfrac{300}{20} = 15</math> cupcakes, and Lacey will frost <math>\dfrac{300}{30} = 10</math> cupcakes. Therefore, working together they will frost <math>15 + 10 = \boxed{\textbf{(D)}\ 25}</math> cupcakes.<br />
<br />
== Solution 3 ==<br />
<br />
Since Cagney frosts <math>3</math> cupcakes a minute, and Lacey frosts <math>2</math> cupcakes a minute, they together frost <math>3+2=5</math> cupcakes a minute. Therefore, in <math>5</math> minutes, they frost <math>5\times5 = 25 \Rightarrow \boxed{\textbf{(D)}}</math><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|before=First Problem|num-a=2}}<br />
{{AMC12 box|year=2012|ab=A|num-b=1|num-a=3}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_14&diff=749182014 AMC 10A Problems/Problem 142016-01-30T20:24:25Z<p>Dipenm: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
The <math>y</math>-intercepts, <math>P</math> and <math>Q</math>, of two perpendicular lines intersecting at the point <math>A(6,8)</math> have a sum of zero. What is the area of <math>\triangle APQ</math>?<br />
<br />
<math> \textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72 </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution 1==<br />
<asy>//Needs refining (hmm I think it's fine --bestwillcui1)<br />
size(12cm);<br />
fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7));<br />
for(int i=-2;i<=8;i+=1)<br />
draw((i,-12)--(i,12),grey);<br />
for(int j=-12;j<=12;j+=1)<br />
draw((-2,j)--(8,j),grey);<br />
draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis<br />
draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis<br />
dot((0,0));<br />
dot((6,8));<br />
draw((-2,10.66667)--(8,7.33333),Arrows);<br />
draw((7.33333,12)--(-0.66667,-12),Arrows);<br />
draw((6,8)--(0,8));<br />
draw((6,8)--(0,0));<br />
draw(rightanglemark((0,10),(6,8),(0,-10),20));<br />
label("$A$",(6,8),NE);<br />
label("$a$", (0,5),W);<br />
label("$a$",(0,-5),W);<br />
label("$a$",(3,4),NW);<br />
label("$P$",(0,10),SW);<br />
label("$Q$",(0,-10),NW);<br />
// wanted to import graph and use xaxis/yaxis but w/e<br />
label("$x$",(9,0),E);<br />
label("$y$",(0,13),N);<br />
</asy><br />
Note that if the <math>y</math>-intercepts have a sum of <math>0</math>, the distance from the origin to each of the intercepts must be the same. Call this distance <math>a</math>. Since the <math>\angle PAQ = 90^\circ</math>, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is <math>\sqrt{6^2+8^2} = 10</math>, this means <math>a=10</math>, and the length of the hypotenuse is <math>2a = 20</math>. Since the <math>x</math>-coordinate of <math>A</math> is the same as the altitude to the hypotenuse, <math>[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}</math>.<br />
<br />
==Solution 2==<br />
<br />
We can let the two lines be <cmath>y=mx+b</cmath> <cmath>y=-\frac{1}{m}x-b</cmath> This is because the lines are perpendicular, hence the <math>m</math> and <math>-\frac{1}{m}</math>, and the sum of the y-intercepts is equal to 0, hence the <math>b, -b</math>. <br />
<br />
Since both lines contain the point <math>(6,8)</math>, we can plug this into the two equations to obtain <cmath>8=6m+b</cmath> and <cmath>8=-6\frac{1}{m}-b</cmath><br />
<br />
Adding the two equations gives <cmath>16=6m+\frac{-6}{m}</cmath> Multiplying by <math>m</math> gives <cmath>16m=6m^2-6</cmath> <cmath>\implies 6m^2-16m-6=0</cmath> <cmath>\implies 3m^2-8m-3=0</cmath> Factoring gives <cmath>(3m+1)(m-3)=0</cmath> <br />
<br />
We can just let <math>m=3</math>, since the two values of <math>m</math> do not affect our solution - one is the slope of one line and the other is the slope of the other line. <br />
<br />
Plugging <math>m=3</math> into one of our original equations, we obtain <cmath>8=6(3)+b</cmath> <cmath>\implies b=8-6(3)=-10</cmath><br />
<br />
Since <math>\bigtriangleup APQ</math> has hypotenuse <math>2|b|=20</math> and the altitude to the hypotenuse is equal to the the x-coordinate of point <math>A</math>, or 6, the area of <math>\bigtriangleup APQ</math> is equal to <cmath>\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\ 60}</cmath><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_12&diff=749172014 AMC 10A Problems/Problem 122016-01-30T20:18:03Z<p>Dipenm: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown What is the area of the shaded region?<br />
<br />
<asy><br />
size(125);<br />
defaultpen(linewidth(0.8));<br />
path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle;<br />
fill(hexagon,grey);<br />
for(int i=0;i<=5;i=i+1)<br />
{<br />
path arc=2*dir(60*i)--arc(2*dir(60*i),1,120+60*i,240+60*i)--cycle;<br />
unfill(arc);<br />
draw(arc);<br />
}<br />
draw(hexagon,linewidth(1.8));</asy><br />
<br />
<math> \textbf{(A)}\ 27\sqrt{3}-9\pi\qquad\textbf{(B)}\ 27\sqrt{3}-6\pi\qquad\textbf{(C)}\ 54\sqrt{3}-18\pi\qquad\textbf{(D)}\ 54\sqrt{3}-12\pi\qquad</math><br />
<br />
<math>\textbf{(E)}\ 108\sqrt{3}-9\pi </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
The area of the hexagon is equal to <math>\dfrac{3(6)^2\sqrt{3}}{2}=54\sqrt{3}</math> by the formula for the area of a hexagon.<br />
<br />
We note that each interior angle of the regular hexagon is <math>120^\circ</math> which means that each sector is <math>\dfrac{1}{3}</math> of the circle it belongs to. The area of each sector is <math>\dfrac{9\pi}{3}=3\pi</math>. The area of all six is <math>6\times 3\pi=18\pi</math>. <br />
<br />
The shaded area is equal to the area of the hexagon minus the sum of the area of all the sectors, which is equal to <math>\boxed{\textbf{(C)}\ 54\sqrt{3}-18\pi}</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_12&diff=749162015 AMC 10A Problems/Problem 122016-01-30T20:02:33Z<p>Dipenm: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Points <math>( \sqrt{\pi} , a)</math> and <math>( \sqrt{\pi} , b)</math> are distinct points on the graph of <math>y^2 + x^4 = 2x^2 y + 1</math>. What is <math>|a-b|</math>?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi} </math><br />
<br />
==Solution #1==<br />
<br />
Since points on the graph make the equation true, substitute <math>\sqrt{\pi}</math> in to the equation and then solve to find <math>a</math> and <math>b</math>.<br />
<br />
<math>y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1</math><br />
<br />
<math>y^2 + \pi^2 = 2\pi y + 1</math><br />
<br />
<math>y^2 - 2\pi y + \pi^2 = 1</math><br />
<br />
<math>(y-\pi)^2 = 1</math><br />
<br />
<math>y-\pi = \pm 1</math><br />
<br />
<math>y = \pi + 1</math> <br />
<br />
<math>y = \pi - 1</math><br />
<br />
There are only two solutions to the equation, so one of them is the value of <math>a</math> and the other is <math>b</math>. The order does not matter because of the absolute value sign.<br />
<br />
<math>| (\pi + 1) - (\pi - 1) | = 2</math><br />
<br />
The answer is <math>\boxed{\textbf{(C) }2}</math><br />
<br />
==Solution #2==<br />
This solution is very related to Solution #1 but just simplifies the problem earlier to make it easier.<br />
<br />
<math>y^2 + x^4 = 2x^2 y + 1</math> can be written as <math>x^4-2x^2y+y^2=1</math>. Recognizing that this is a binomial square, simplify this to <math>(x^2-y)^2=1</math>. This gives us two equations:<br />
<br />
<math>x^2-y=1</math> and <math>x^2-y=-1</math>.<br />
<br />
One of these <math>y</math>'s is <math>a</math> and one is <math>b</math>. Substituting <math>\sqrt{\pi}</math> for <math>x</math>, we get <math>a=\pi+1</math> and <math>b=\pi-1</math>. <br />
<br />
So, <math>|a-b|=|(\pi+1)-(\pi-1)|=2</math>.<br />
<br />
The answer is <math>\boxed{\textbf{(C) }2}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Algebra Problems]]</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_12&diff=749152015 AMC 10A Problems/Problem 122016-01-30T20:02:03Z<p>Dipenm: /* Alternate Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Points <math>( \sqrt{\pi} , a)</math> and <math>( \sqrt{\pi} , b)</math> are distinct points on the graph of <math>y^2 + x^4 = 2x^2 y + 1</math>. What is <math>|a-b|</math>?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi} </math><br />
<br />
==Solution==<br />
<br />
Since points on the graph make the equation true, substitute <math>\sqrt{\pi}</math> in to the equation and then solve to find <math>a</math> and <math>b</math>.<br />
<br />
<math>y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1</math><br />
<br />
<math>y^2 + \pi^2 = 2\pi y + 1</math><br />
<br />
<math>y^2 - 2\pi y + \pi^2 = 1</math><br />
<br />
<math>(y-\pi)^2 = 1</math><br />
<br />
<math>y-\pi = \pm 1</math><br />
<br />
<math>y = \pi + 1</math> <br />
<br />
<math>y = \pi - 1</math><br />
<br />
There are only two solutions to the equation, so one of them is the value of <math>a</math> and the other is <math>b</math>. The order does not matter because of the absolute value sign.<br />
<br />
<math>| (\pi + 1) - (\pi - 1) | = 2</math><br />
<br />
The answer is <math>\boxed{\textbf{(C) }2}</math><br />
<br />
==Solution #2==<br />
This solution is very related to Solution #1 but just simplifies the problem earlier to make it easier.<br />
<br />
<math>y^2 + x^4 = 2x^2 y + 1</math> can be written as <math>x^4-2x^2y+y^2=1</math>. Recognizing that this is a binomial square, simplify this to <math>(x^2-y)^2=1</math>. This gives us two equations:<br />
<br />
<math>x^2-y=1</math> and <math>x^2-y=-1</math>.<br />
<br />
One of these <math>y</math>'s is <math>a</math> and one is <math>b</math>. Substituting <math>\sqrt{\pi}</math> for <math>x</math>, we get <math>a=\pi+1</math> and <math>b=\pi-1</math>. <br />
<br />
So, <math>|a-b|=|(\pi+1)-(\pi-1)|=2</math>.<br />
<br />
The answer is <math>\boxed{\textbf{(C) }2}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Algebra Problems]]</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_12&diff=749142015 AMC 10A Problems/Problem 122016-01-30T19:39:00Z<p>Dipenm: /* Alternate Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Points <math>( \sqrt{\pi} , a)</math> and <math>( \sqrt{\pi} , b)</math> are distinct points on the graph of <math>y^2 + x^4 = 2x^2 y + 1</math>. What is <math>|a-b|</math>?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi} </math><br />
<br />
==Solution==<br />
<br />
Since points on the graph make the equation true, substitute <math>\sqrt{\pi}</math> in to the equation and then solve to find <math>a</math> and <math>b</math>.<br />
<br />
<math>y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1</math><br />
<br />
<math>y^2 + \pi^2 = 2\pi y + 1</math><br />
<br />
<math>y^2 - 2\pi y + \pi^2 = 1</math><br />
<br />
<math>(y-\pi)^2 = 1</math><br />
<br />
<math>y-\pi = \pm 1</math><br />
<br />
<math>y = \pi + 1</math> <br />
<br />
<math>y = \pi - 1</math><br />
<br />
There are only two solutions to the equation, so one of them is the value of <math>a</math> and the other is <math>b</math>. The order does not matter because of the absolute value sign.<br />
<br />
<math>| (\pi + 1) - (\pi - 1) | = 2</math><br />
<br />
The answer is <math>\boxed{\textbf{(C) }2}</math><br />
<br />
==Alternate Solution==<br />
<br />
<math>y^2 + x^4 = 2x^2 y + 1</math> can be written as <math>x^4-2x^2y+y^2=1</math>. Recognizing that this is a binomial square, simplify this to <math>(x^2-y)^2=1</math>. This gives us two equations:<br />
<br />
<math>x^2-y=1</math> and <math>x^2-y=-1</math>.<br />
<br />
One of these <math>y</math>'s is <math>a</math> and one is <math>b</math>. Substituting <math>\sqrt{\pi}</math> for <math>x</math>, we get <math>a=\pi+1</math> and <math>b=\pi-1</math>. <br />
<br />
So, <math>|a-b|=|(\pi+1)-(\pi-1)|=2</math>.<br />
<br />
The answer is <math>\boxed{\textbf{(C) }2}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Algebra Problems]]</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_12&diff=749132015 AMC 10A Problems/Problem 122016-01-30T19:36:51Z<p>Dipenm: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Points <math>( \sqrt{\pi} , a)</math> and <math>( \sqrt{\pi} , b)</math> are distinct points on the graph of <math>y^2 + x^4 = 2x^2 y + 1</math>. What is <math>|a-b|</math>?<br />
<br />
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi} </math><br />
<br />
==Solution==<br />
<br />
Since points on the graph make the equation true, substitute <math>\sqrt{\pi}</math> in to the equation and then solve to find <math>a</math> and <math>b</math>.<br />
<br />
<math>y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1</math><br />
<br />
<math>y^2 + \pi^2 = 2\pi y + 1</math><br />
<br />
<math>y^2 - 2\pi y + \pi^2 = 1</math><br />
<br />
<math>(y-\pi)^2 = 1</math><br />
<br />
<math>y-\pi = \pm 1</math><br />
<br />
<math>y = \pi + 1</math> <br />
<br />
<math>y = \pi - 1</math><br />
<br />
There are only two solutions to the equation, so one of them is the value of <math>a</math> and the other is <math>b</math>. The order does not matter because of the absolute value sign.<br />
<br />
<math>| (\pi + 1) - (\pi - 1) | = 2</math><br />
<br />
The answer is <math>\boxed{\textbf{(C) }2}</math><br />
<br />
==Alternate Solution==<br />
<br />
<math>y^2 + x^4 = 2x^2 y + 1</math> can be written as <math>x^4-2x^2y+y^2=1</math>. Recognizing that this is a binomial square, simplify this to <math>(x^2-y)^2=1</math>. This gives us two equations:<br />
<br />
<math>x^2-y=1</math> and <math>x^2-y=-1</math>.<br />
<br />
One of these <math>y</math>s is <math>a</math> and one is <math>b</math>.Substituting <math>\sqrt{\pi}</math> for <math>x</math>, we get <math>a=\pi+1</math> and <math>b=\pi-1</math>. So, <math>|a-b|=|(\pi+1)-(\pi-1)|=2</math>.<br />
<br />
The answer is <math>\boxed{\textbf{(C) }2}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Algebra Problems]]</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_17&diff=678102014 AMC 10B Problems/Problem 172015-02-12T03:58:15Z<p>Dipenm: </p>
<hr />
<div>==Problem 17==<br />
What is the greatest power of <math>2</math> that is a factor of <math>10^{1002} - 4^{501}</math>?<br />
<br />
<math>\textbf{(A) } 2^{1002} \qquad\textbf{(B) } 2^{1003} \qquad\textbf{(C) } 2^{1004} \qquad\textbf{(D) } 2^{1005} \qquad\textbf{(E) }2^{1006}</math><br />
<br />
==Solution 1==<br />
<br />
We begin by factoring the <math>2^{1002}</math> out. This leaves us with <math>5^{1002} - 1</math>. <br />
<br />
We factor the difference of squares, leaving us with <math>(5^{501} - 1)(5^{501} + 1)</math>. We note that all even powers of 5 more than two end in ...<math>625</math>. Also, all odd powers of five more than 2 end in ...<math>125</math>. Thus, <math>(5^{501} + 1)</math> would end in ...<math>126</math> and thus would contribute one power of two to the answer, but not more. <br />
<br />
We can continue to factor <math>(5^{501} - 1)</math> as a difference of cubes, leaving us with <math>(5^{167} - 1)</math> times an odd number. <math>(5^{167} - 1)</math> ends in ...<math>124</math>, contributing two powers of two to the final result.<br />
<br />
Adding these extra <math>3</math> powers of two to the original <math>1002</math> factored out, we obtain the final answer of <math>\textbf{(D) } 2^{1005}</math>.<br />
<br />
==Solution 2==<br />
<br />
First, we can write the expression in a more primitive form which will allow us to start factoring.<br />
<cmath>10^{1002} - 4^{501} = 2^{1002} \cdot 5^{1002} - 2^{1002}</cmath><br />
Now, we can factor out <math>2^{1002}</math>. This leaves us with <math>5^{1002} - 1</math>. Call this number <math>N</math>. Thus, our final answer will be <math>2^{1002+k}</math>, where <math>k</math> is the largest power of <math>2</math> that divides <math>N</math>. Now we can consider <math>N \pmod{16}</math>, since <math>k \le 4</math> by the answer choices. <br />
<br />
Note that<br />
<cmath>\begin{align*} 5^1 &\equiv 5 \pmod{16} \\ 5^2 &\equiv 9 \pmod{16} \\ 5^3 &\equiv 13 \pmod{16} \\ 5^4 &\equiv 1 \pmod{16} \\ 5^5 &\equiv 5 \pmod{16} \\ &\: \: \qquad \vdots \end{align*}</cmath><br />
The powers of <math>5</math> cycle in <math>\mod{16}</math> with a period of <math>4</math>. Thus, <cmath>5^{1002} \equiv 5^2 \equiv 9 \pmod{16} \implies 5^{1002} - 1 \equiv 8 \pmod{16}</cmath><br />
This means that <math>N</math> is divisible by <math>8= 2^3</math> but not <math>16 = 2^4</math>, so <math>k = 3</math> and our answer is <math>2^{1002 + 3} =\boxed{\textbf{(D)}\: 2^{1005}}</math>.<br />
<br />
<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=Asymptote:_Basics&diff=62129Asymptote: Basics2014-05-29T16:29:06Z<p>Dipenm: /* Syntax */</p>
<hr />
<div>{{Asymptote}}<br />
<br />
== Syntax ==<br />
In Asymptote, every command should be followed by a semicolon (;) and then an asy tag [asy]. The whole command should end with an end asy tag [/asy] This tells Asymptote how to separate one command from the next. For example, the command <tt>draw(A--B)</tt> draws a line segment from point <math>A</math> to <math>B</math> given as coordinate pairs. Thus, if you wanted to draw two line segments, say one of them from (0,0) to (50,50) (in units of PostScript points - see below for the explanation of units and size) and the other from (50,0) to (0,50), you can create the following document:<br />
draw((0,0)--(50,50));<br />
draw((0,50)--(50,0));<br />
<br />
The two commands do not need to be on separate lines; it is the semicolon that separates them. However, putting the commands on separate lines does not change the output, so it is often useful to separate your commands in this way in order to organize your code. Whitespaces before and after commands are also not read by Asymptote, so any line can be indented as far as desired for clarity's sake.<br />
<br />
To write comments in your code that do not affect the output at all, simply start a line with two forward slashes: <tt>//</tt> as in<br />
// The following line is drawn from (0,0) to (50,50)<br />
draw((0,0)--(50,50));<br />
<br />
==Variables and Data Types==<br />
Asymptote parses your code into substrings which have a certain '''data type''', for example a ''real'' (like <math>1.5</math>) or a ''pair'' (like <math>(2,3)</math>). Each new variable that you declare must be declared as a data type that Asymptote recognizes, by the command <tt>[datatype] [variable];</tt>. For example, if you wanted to declare the variable <math>n</math> to have type integer, you can use the command<br />
int n;<br />
<br />
After it is declared, you can store a specific value in a variable using the <math>=</math> symbol, as in<br />
n=3;<br />
<br />
These two commands can be abbreviated by the single command <tt>int n=3;</tt>, and several integers can also be declared at once (<tt>int m,n,d;</tt>) or even many declarations and assignations at once: <tt>int a,b=2,c,d=5;</tt>.<br />
<br />
As another example of variable declaration, consider the picture of the X from the Syntax section above. The same picture can be made as follows:<br />
pair A,B,C,D;<br />
A=(0,0);<br />
B=(50,50);<br />
C=(0,50);<br />
D=(50,0);<br />
draw(A--B);<br />
draw(C--D);<br />
In this particular example, variable declarations made the code longer, but as you will see, declaring variables will significantly clean up your code in messier diagrams.<br />
<br />
The most commonly used data types in Asymptote are given in the following table:<br />
<center>[[Image:Table1.gif]]</center><br />
<br />
==Size and Unitsize==<br />
Asymptote is a primarily coordinate-based graphics language. Each point is a pair <math>(a,b)</math> where <math>a</math> is the <math>x</math>-coordinate and <math>b</math> is the <math>y</math>-coordinate.<br />
<br />
However, there are many ways to choose a Cartesian coordinate system for the plane; one must pick the placement of origin and the scale on each of the <math>x</math>- and <math>y</math>-axes. Asymptote will place your image in the center of your output page after it is drawn, so placement of origin is actually irrelevant. By default, the unit length in both the <math>x</math> and <math>y</math> directions is the PostScript bigpoint, which has length <math>1/72</math> inches. Thus, if you do not change the scaling on the picture, the points <math>(0,0)</math> and <math>(72,0)</math> are exactly one inch apart when drawn in Asymptote. However, drawing in bigpoints is inconvenient if you wish to draw a figure that is exactly 3cm wide.<br />
<br />
The function <tt>unitsize</tt> can be used to specify the unit length for your picture. This function takes up to 3 arguments: the picture you want to scale the axes for (if this isn't specified, it defaults to <tt>currentpicture</tt>, the picture you are drawing on), the unit length in the x direction, and the unit length in the y direction. If only one real argument is given, both the x and y unit sizes are set to this number. Thus the command<br />
unitsize(72);<br />
will tell Asymptote that from now on, your unit length is <math>1</math> inch. Be careful when you are redefining your unit length - now that unitsize is set to <math>72</math>, the points <math>(0,0)</math> and <math>(72,0)</math> are actually <math>72</math> inches apart!<br />
<br />
Asymptote has the built-in constants <tt>pt</tt> (1/72.27 inches), <tt>inch</tt>, <tt>cm</tt>, and <tt>mm</tt> for convenience when defining lengths, so the above command can also be stated<br />
unitsize(1inch);<br />
<br />
The other useful function is <tt>size</tt>, which specifies the exact width and height of the box that your picture (if unspecified as a first argument, this will again default to <tt>currentpicture</tt>) will be fit into. If only one number is given, both the width and the height will be set to this number. For example, the command<br />
size(5cm,5cm);<br />
or just<br />
size(5cm);<br />
will fit the diagram to a 5cm x 5cm box regardless of the specified unitsizes.<br />
<br />
As an example, make an Asymptote document containing the following two lines:<br />
unitsize(2inch); <br />
draw(unitsquare);<br />
and see what happens as you change the <math>2</math> inch size to several other values.<br />
<br />
See also: [[Asymptote: Drawing]]<br />
<br />
[[Asymptote: Reference | Next: Reference]]</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=MATHCOUNTS&diff=59853MATHCOUNTS2014-02-18T23:23:27Z<p>Dipenm: </p>
<hr />
<div>'''MathCounts''' is a large national [[mathematics competition]] and [[mathematics coaching]] program that has served millions of middle school students since 1984. Sponsored by the [[CNA Foundation]], [[National Society of Professional Engineers]], the [[National Council of Teachers of Mathematics]], and others, the focus of MATHCOUNTS is on [[mathematical problem solving]]. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.<br />
<br />
== MATHCOUNTS Curriculum ==<br />
MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.<br />
<br />
Before 1990, MATHCOUNTS chose particular areas of mathematics to highlight each year before changing the focus of the competition more broadly to problem solving.<br />
<br />
==Past Winners==<br />
* 1984: Michael Edwards, Texas<br />
* 1985: Timothy Kokesh, Oklahoma<br />
* 1986: Brian David Ewald, Florida<br />
* 1987: Russell Mann, Tennessee<br />
* 1988: Andrew Schultz, Illinois<br />
* 1989: Albert Kurz, Pennsylvania<br />
* 1990: Brian Jenkins, Arkansas<br />
* 1991: Jonathan L. Weinstein, Massachusetts<br />
* 1992: Andrei C. Gnepp, Ohio<br />
* 1993: Carleton Bosley, Kansas<br />
* 1994: William O. Engel, Illinois<br />
* 1995: Richard Reifsnyder, Kentucky<br />
* 1996: Alexander Schwartz, Pennsylvania<br />
* 1997: Zhihao Liu, Wisconsin<br />
* 1998: Ricky Liu, Massachusetts<br />
* 1999: Po-Ru Loh, Wisconsin<br />
* 2000: Ruozhou Jia, Illinois<br />
* 2001: Ryan Ko, New Jersey<br />
* 2002: Albert Ni, Illinois<br />
* 2003: Adam Hesterberg, Washington<br />
* 2004: Gregory Gauthier, Illinois<br />
* 2005: Neal Wu, Louisiana<br />
* 2006: Daesun Yim, New Jersey<br />
* 2007: Kevin Chen, Texas<br />
* 2008: Darryl Wu, Washington<br />
* 2009: Bobby Shen, Texas<br />
* 2010: Mark Sellke, Indiana<br />
* 2011: Scott Wu, Louisiana<br />
* 2012: Chad Qian, Indiana<br />
* 2013: Alec Sun, Massachusetts<br />
<br />
== Past State Team Winners ==<br />
* 1984: Virginia<br />
* 1985: Florida<br />
* 1986: California<br />
* 1987: New York<br />
* 1988: New York<br />
* 1989: North Carolina<br />
* 1990: Ohio<br />
* 1991: Alabama<br />
* 1992: California<br />
* 1993: Kansas<br />
* 1994: Pennsylvania<br />
* 1995: Indiana<br />
* 1996: Wisconsin<br />
* 1997: Massachusetts<br />
* 1998: Wisconsin<br />
* 1999: Massachusetts<br />
* 2000: California<br />
* 2001: Virginia<br />
* 2002: California<br />
* 2003: California<br />
* 2004: Illinois<br />
* 2005: Texas<br />
* 2006: Virginia<br />
* 2007: Texas<br />
* 2008: Texas<br />
* 2009: Texas<br />
* 2010: California<br />
* 2011: California<br />
* 2012: Massachusetts<br />
* 2013: Massachusetts<br />
<br />
== MATHCOUNTS Competition Structure ==<br />
<br />
=== Sprint Round ===<br />
<br />
30 problems in 40 minutes. This round is generally made up questions ranging from relatively easy to relatively difficult. Some of the difficult problems are only difficult because calculators are not allowed in this round. The questions increase in difficulty as the test progresses, so the 30th question is likely to be harder than the first. Geometry problems in this round can be very challenging. The AoPS books will help greatly in training for this round, or any round.<br />
<br />
=== Target Round ===<br />
8 problems given 2 at a time. Each set of two problems is given six minutes. Students may not go back to previous rounds (or forwards to future rounds) even if they finish before time is called. Students may use calculators. The AoPS books will help greatly in training for this round, or any round.<br />
<br />
=== Team Round ===<br />
<br />
10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions).<br />
<br />
=== Countdown Round ===<br />
High scoring individuals compete head-to-head until a champion is crowned. People compete from off a screen taking 45 seconds or less to finish the problem. The Countdown round is run differently in various different chapter, state, and national competitions. In the national competitions, it is the round that determines the champion. Calculators are not allowed.<br />
<br />
<br />
====Chapter and State Competitions====<br />
<br />
In the chapter and state competitions, the countdown round is not mandatory. However, if it is deemed official by the chapter or state, the following format must be used:<br />
<br />
*The 10th place written finisher competes against the 9th place written finisher. A problem is displayed, and both competitors have 45 seconds to answer the question, and the first competitor to correctly answer the question receives one point. The person who gets the most correct out of three questions (not necessarily two out of three) is the winner.<br />
<br />
*The winner of the first round goes up against the 8th place finisher.<br />
<br />
*The winner of the second round goes up against the 7th place finisher.<br />
<br />
This process is continued until the countdown round reaches the top four written competitors. Starting then, the first person to get three questions correct wins (as opposed to the best-out-of-three rule).<br />
<br />
If the countdown round is unofficial, any format may be used. Single-elimination bracket-style tournaments are common.<br />
<br />
====National Competition====<br />
<br />
At the national competition, there are some structural changes to the countdown round. The top 12 (not the top 10) written finishers make it to the countdown round, and the format is changed from a ladder competition to a single elimination tournament where the top four written competitors get a bye. This setup makes it far more likely for a 12th place finisher to become champion, and it makes it less likely for a first place written finisher to become champion, equalizing the field. But even then, a 12th place written competitor will have less of a chance to become champion than the top 4, because the top 4 get a bye. Until the semi-finals, the scoring is best out of five advances.<br />
<br />
At the first round and the second round, the person to correctly answer the most out of 5 questions wins. However, at the semifinals, the rules slightly change&mdash;the first person to correctly answer four questions wins.<br />
<br />
=== Ciphering Round ===<br />
In some states, (most notably Florida) there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representative to go against all representatives from the other teams. A problem is shown on a screen and students work fast to answer the problem. The students give their answer and after 45 seconds the answer is shown and the answers are checked to see if they are right. The fastest correct answer gets five points, the next fastest gets 4, etc. There are 4 questions per individual and teams send up 4 people. A perfect score is then 80. Often times the questions take clever reading skills. For example, one question was "How much dirt is in a 3 ft by 3 ft by 4 ft hole?" The answer was 0 because there is no dirt in a hole.<br />
<br />
=== Masters Round ===<br />
Top students give in-depth explanations to challenging problems. This round is optional at the state level competition and is mandatory at the national competition (up to 2011). At nationals the top two on the written and countdown participate. <br />
In 2012, it was replaced by the Reel Math Challenge (now called the Math Video Challenge).<br />
<br />
=== Scoring and Ranking ===<br />
An individual's score is their total number of correct sprint round answers plus 2 times their total number of correct target round answers. This total is out of a maximum of <math>30 + 2(8) = 46</math> points.<br />
<br />
A team's score is the individual scores of its members divided by 4 plus 2 points for every correct team round answer, making a team's maximum possible score 66 points. Therefore, it is possible to win with a relatively low team score and a phenomenal individual score, as the team score is only roughly 30% of the total team score. Note that when there are less than four members the score will become less.<br />
<br />
== MATHCOUNTS Competition Levels ==<br />
=== School Competition ===<br />
Students vie for the chance to make their school teams. Problems at this level are generally the easiest and most basic in curriculum.<br />
<br />
=== Chapter Competition ===<br />
Chapter competitions serve as a selection filter for state competitions. A few states don't need to host chapter competitions due to a small population size.<br />
<br />
=== State Competition ===<br />
The top 4 students in each state form the state team for the national competition. The coach of the top school team at the state level is invited to coach the state team at the national competition. Interestingly, the coach of a state team is not necessarily the coach of any of the state's team members.<br />
<br />
=== National Competition ===<br />
==== National Competition Sites ====<br />
For many years, the National MATHCOUNTS competition was held in Washington, D.C. More recently, the competition has changed venues often.<br />
<br />
* The 2013 competition was held in Washington, D.C.<br />
* The 2012 competition was held in Orlando, Florida.<br />
* The 2011 competition was held in Washington, D.C.<br />
* The 2009 and 2010 competitions were held in Orlando, Florida.<br />
* The 2008 competition was held in Denver, Colorado.<br />
* The 2007 competition was held in Fort Worth, Texas.<br />
* The 2006 competition was held in Arlington, Virginia.<br />
* The 2005 competition was held in Detroit, Michigan.<br />
* The 2004 competition was held in Washington, D.C.<br />
* The 2002 and 2003 competitions were held in Chicago, Illinois.<br />
<br />
==== Rewards ====<br />
<br />
Every competitor at the national competition receives a graphing calculator that varies by year - for example, in 2006 it was a TI-84 Plus Silver Edition with the MATHCOUNTS logo on the back. In 2007, MATHCOUNTS took the logo off. In 2008 to 2012, they gave TI-<math>n</math>spires to everyone. They also give out a laptop and an 8000 dollar scholarship to the countdown winner and the written winner. They also gave 6000 dollars to 2nd place, and 4000 dollars to the semifinalists. (The written round winner also got 8000 dollars)<br />
<br />
== MATHCOUNTS Resources ==<br />
=== MATHCOUNTS Books ===<br />
* [http://www.artofproblemsolving.com/Books/AoPS_B_CP_MC.php MATHCOUNTS books] at the [http://www.artofproblemsolving.com/Books/AoPS_B_About.php AoPS Bookstore]<br />
* [[Art of Problem Solving]]'s [http://www.artofproblemsolving.com/Books/AoPS_B_Rec_Middle.php Introductory subject textbooks] are ideal for students preparing for MATHCOUNTS.<br />
<br />
=== MATHCOUNTS Classes ===<br />
* [[Art of Problem Solving]] hosts [http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php#mc MATHCOUNTS preparation classes].<br />
* [[Art of Problem Solving]] hosts many free MATHCOUNTS [[Math Jams]]. [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jam Schedule]. [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php Math Jam Transcript Archive].<br />
<br />
=== MATHCOUNTS Online ===<br />
* [http://www.mathcounts.org MATHCOUNTS Homepage]<br />
* [[Art of Problem Solving]] hosts a large [http://www.artofproblemsolving.com/Forum/index.php?f=132 MATHCOUNTS Forum] as well as a private [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23209 MATHCOUNTS Coaches Forum].<br />
* [http://mathcounts.saab.org/ Elias Saab's MATHCOUNTS Preparation Homepage]<br />
* [http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html The MATHCOUNTS Bible According to Mr. Diaz]<br />
*[http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php/ Building a Successful MATHCOUNTS Program] by [[Jeff Boyd]], who coached the 2005, 2007, and 2008 National Champion [[Texas MathCounts]] team.<br />
<br />
== What comes after MATHCOUNTS? ==<br />
<br />
Give the following competitions a try and take a look at the [[List of United States high school mathematics competitions]].<br />
* [[American Mathematics Competitions]]<br />
* [[American Regions Math League]]<br />
* [[Mandelbrot Competition]]<br />
* [[Mu Alpha Theta]]<br />
<br />
[[Category:Mathematics competitions]]<br />
<br />
== See also... ==<br />
* [[List of national MATHCOUNTS teams]]<br />
* [[MATHCOUNTS historical results]]<br />
* [[Mathematics competition resources]]<br />
* [[Math contest books]]<br />
* [[Math books]]<br />
* [[List of United States middle school mathematics competitions]]<br />
* [[List of United States high school mathematics competitions]]<br />
* [http://www.mathcounts.org/webarticles/anmviewer.asp?a=921&z=71 2006 MATHCOUNTS Countdown Video]</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=MATHCOUNTS&diff=58689MATHCOUNTS2014-01-12T02:10:40Z<p>Dipenm: /* See also */</p>
<hr />
<div>'''MATHCOUNTS''' is a large national [[mathematics competition]] and [[mathematics coaching]] program that has served millions of middle school students since 1984. Sponsored by the [[CNA Foundation]], [[National Society of Professional Engineers]], the [[National Council of Teachers of Mathematics]], and others, the focus of MATHCOUNTS is on [[mathematical problem solving]]. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.<br />
<br />
== MATHCOUNTS Curriculum ==<br />
MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.<br />
<br />
Before 1990, MATHCOUNTS chose particular areas of mathematics to highlight each year before changing the focus of the competition more broadly to problem solving.<br />
<br />
==Past Winners==<br />
* 1984: Michael Edwards, Texas<br />
* 1985: Timothy Kokesh, Oklahoma<br />
* 1986: Brian David Ewald, Florida<br />
* 1987: Russell Mann, Tennessee<br />
* 1988: Andrew Schultz, Illinois<br />
* 1989: Albert Kurz, Pennsylvania<br />
* 1990: Brian Jenkins, Arkansas<br />
* 1991: Jonathan L. Weinstein, Massachusetts<br />
* 1992: Andrei C. Gnepp, Ohio<br />
* 1993: Carleton Bosley, Kansas<br />
* 1994: William O. Engel, Illinois<br />
* 1995: Richard Reifsnyder, Kentucky<br />
* 1996: Alexander Schwartz, Pennsylvania<br />
* 1997: Zhihao Liu, Wisconsin<br />
* 1998: Ricky Liu, Massachusetts<br />
* 1999: Po-Ru Loh, Wisconsin<br />
* 2000: Ruozhou Jia, Illinois<br />
* 2001: Ryan Ko, New Jersey<br />
* 2002: Albert Ni, Illinois<br />
* 2003: Adam Hesterberg, Washington<br />
* 2004: Gregory Gauthier, Illinois<br />
* 2005: Neal Wu, Louisiana<br />
* 2006: Daesun Yim, New Jersey<br />
* 2007: Kevin Chen, Texas<br />
* 2008: Darryl Wu, Washington<br />
* 2009: Bobby Shen, Texas<br />
* 2010: Mark Sellke, Indiana<br />
* 2011: Scott Wu, Louisiana<br />
* 2012: Chad Qian, Indiana<br />
* 2013: Alec Sun, Massachusetts<br />
<br />
== Past State Team Winners ==<br />
* 1984: Virginia<br />
* 1985: Florida<br />
* 1986: California<br />
* 1987: New York<br />
* 1988: New York<br />
* 1989: North Carolina<br />
* 1990: Ohio<br />
* 1991: Alabama<br />
* 1992: California<br />
* 1993: Kansas<br />
* 1994: Pennsylvania<br />
* 1995: Indiana<br />
* 1996: Wisconsin<br />
* 1997: Massachusetts<br />
* 1998: Wisconsin<br />
* 1999: Massachusetts<br />
* 2000: California<br />
* 2001: Virginia<br />
* 2002: California<br />
* 2003: California<br />
* 2004: Illinois<br />
* 2005: Texas<br />
* 2006: Virginia<br />
* 2007: Texas<br />
* 2008: Texas<br />
* 2009: Texas<br />
* 2010: California<br />
* 2011: California<br />
* 2012: Massachusetts<br />
* 2013: Massachusetts<br />
<br />
== MATHCOUNTS Competition Structure ==<br />
<br />
=== Sprint Round ===<br />
<br />
30 problems in 40 minutes. This round is generally made up questions ranging from relatively easy to relatively difficult. Some of the difficult problems are only difficult because calculators are not allowed in this round.<br />
<br />
=== Target Round ===<br />
8 problems given 2 at a time. Each set of two problems is given six minutes. Students may not go back to previous rounds (or forwards to future rounds) even if they finish before time is called. Students may use calculators.<br />
<br />
=== Team Round ===<br />
<br />
10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions).<br />
<br />
=== Countdown Round ===<br />
High scoring individuals compete head-to-head until a champion is crowned. People compete from off a screen taking 45 seconds or less to finish the problem. The Countdown round is run differently in various different chapter, state, and national competitions. In the national competitions, it is the round that determines the champion. Calculators are not allowed.<br />
<br />
<br />
====Chapter and State Competitions====<br />
<br />
In the chapter and state competitions, the countdown round is not mandatory. However, if it is deemed official by the chapter or state, the following format must be used:<br />
<br />
*The 10th place written finisher competes against the 9th place written finisher. A problem is displayed, and both competitors have 45 seconds to answer the question, and the first competitor to correctly answer the question receives one point. The person who gets the most correct out of three questions (not necessarily two out of three) is the winner.<br />
<br />
*The winner of the first round goes up against the 8th place finisher.<br />
<br />
*The winner of the second round goes up against the 7th place finisher.<br />
<br />
This process is continued until the countdown round reaches the top four written competitors. Starting then, the first person to get three questions correct wins (as opposed to the best-out-of-three rule).<br />
<br />
If the countdown round is unofficial, any format may be used. Single-elimination bracket-style tournaments are common.<br />
<br />
====National Competition====<br />
<br />
At the national competition, there are some structural changes to the countdown round. The top 12 (not the top 10) written finishers make it to the countdown round, and the format is changed from a ladder competition to a single elimination tournament where the top four written competitors get a bye. This setup makes it far more likely for a 12th place finisher to become champion, and it makes it less likely for a first place written finisher to become champion, equalizing the field. But even then, a 12th place written competitor will have less of a chance to become champion than the top 4, because the top 4 get a bye. Until the semi-finals, the scoring is best out of five advances.<br />
<br />
At the first round and the second round, the person to correctly answer the most out of 5 questions wins. However, at the semifinals, the rules slightly change&mdash;the first person to correctly answer four questions wins.<br />
<br />
=== Ciphering Round ===<br />
In some states, (most notably Florida) there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representative to go against all representatives from the other teams. A problem is shown on a screen and students work fast to answer the problem. The students give their answer and after 45 seconds the answer is shown and the answers are checked to see if they are right. The fastest correct answer gets five points, the next fastest gets 4, etc. There are 4 questions per individual and teams send up 4 people. A perfect score is then 80. Often times the questions take clever reading skills. For example, one question was "How much dirt is in a 3 ft by 3 ft by 4 ft hole?" The answer was 0 because there is no dirt in a hole.<br />
<br />
=== Masters Round ===<br />
Top students give in-depth explanations to challenging problems. This round is optional at the state level competition and is mandatory at the national competition (up to 2011). At nationals the top two on the written and countdown participate. <br />
In 2012, it was replaced by the Reel Math Challenge.<br />
<br />
=== Scoring and Ranking ===<br />
An individual's score is their total number of correct sprint round answers plus 2 times their total number of correct target round answers. This total is out of a maximum of <math>30 + 2(8) = 46</math> points.<br />
<br />
A team's score is the individual scores of its members divided by 4 plus 2 points for every correct team round answer, making a team's maximum possible score 66 points. Therefore, it is possible to win with a relatively low team score and a phenomenal individual score, as the team score is only roughly 30% of the total team score. Note that when there are less than four members the score will become less.<br />
<br />
== MATHCOUNTS Competition Levels ==<br />
=== School Competition ===<br />
Students vie for the chance to make their school teams. Problems at this level are generally the easiest and most basic in curriculum.<br />
<br />
=== Chapter Competition ===<br />
Chapter competitions serve as a selection filter for state competitions. A few states don't need to host chapter competitions due to a small population size.<br />
<br />
=== State Competition ===<br />
The top 4 students in each state form the state team for the national competition. The coach of the top school team at the state level is invited to coach the state team at the national competition. Interestingly, the coach of a state team is not necessarily the coach of any of the state's team members.<br />
<br />
=== National Competition ===<br />
==== National Competition Sites ====<br />
For many years, the National MATHCOUNTS competition was held in Washington, D.C. More recently, the competition has changed venues often.<br />
<br />
* The 2013 competition was held in Washington, D.C.<br />
* The 2012 competition was held in Orlando, Florida.<br />
* The 2011 competition was held in Washington, D.C.<br />
* The 2009 and 2010 competitions were held in Orlando, Florida.<br />
* The 2008 competition was held in Denver, Colorado.<br />
* The 2007 competition was held in Fort Worth, Texas.<br />
* The 2006 competition was held in Arlington, Virginia.<br />
* The 2005 competition was held in Detroit, Michigan.<br />
* The 2004 competition was held in Washington, D.C.<br />
* The 2002 and 2003 competitions were held in Chicago, Illinois.<br />
<br />
==== Rewards ====<br />
<br />
Every competitor at the national competition receives a graphing calculator that varies by year - for example, in 2006 it was a TI-84 Plus Silver Edition with the MATHCOUNTS logo on the back. In 2007, MATHCOUNTS took the logo off. In 2008 to 2012, they gave TI-<math>n</math>spires to everyone. They also give out a laptop and an 8000 dollar scholarship to the countdown winner and the written winner. They also gave 6000 dollars to 2nd place, and 4000 dollars to the semifinalists. (The written round winner also got 8000 dollars)<br />
<br />
== MATHCOUNTS Resources ==<br />
=== MATHCOUNTS Books ===<br />
* [http://www.artofproblemsolving.com/Books/AoPS_B_CP_MC.php MATHCOUNTS books] at the [http://www.artofproblemsolving.com/Books/AoPS_B_About.php AoPS Bookstore]<br />
* [[Art of Problem Solving]]'s [http://www.artofproblemsolving.com/Books/AoPS_B_Rec_Middle.php Introductory subject textbooks] are ideal for students preparing for MATHCOUNTS.<br />
<br />
=== MATHCOUNTS Classes ===<br />
* [[Art of Problem Solving]] hosts [http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php#mc MATHCOUNTS preparation classes].<br />
* [[Art of Problem Solving]] hosts many free MATHCOUNTS [[Math Jams]]. [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jam Schedule]. [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php Math Jam Transcript Archive].<br />
<br />
=== MATHCOUNTS Online ===<br />
* [http://www.mathcounts.org MATHCOUNTS Homepage]<br />
* [[Art of Problem Solving]] hosts a large [http://www.artofproblemsolving.com/Forum/index.php?f=132 MATHCOUNTS Forum] as well as a private [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23209 MATHCOUNTS Coaches Forum].<br />
* [http://mathcounts.saab.org/ Elias Saab's MATHCOUNTS Preparation Homepage]<br />
* [http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html The MATHCOUNTS Bible According to Mr. Diaz]<br />
*[http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php/ Building a Successful MATHCOUNTS Program] by [[Jeff Boyd]], who coached the 2005, 2007, and 2008 National Champion [[Texas MathCounts]] team.<br />
<br />
== What comes after MATHCOUNTS? ==<br />
<br />
Give the following competitions a try and take a look at the [[List of United States high school mathematics competitions]].<br />
* [[American Mathematics Competitions]]<br />
* [[American Regions Math League]]<br />
* [[Mandelbrot Competition]]<br />
* [[Mu Alpha Theta]]<br />
<br />
[[Category:Mathematics competitions]]<br />
<br />
== See also... ==<br />
* [[List of national MATHCOUNTS teams]]<br />
* [[MATHCOUNTS historical results]]<br />
* [[Mathematics competition resources]]<br />
* [[Math contest books]]<br />
* [[Math books]]<br />
* [[List of United States middle school mathematics competitions]]<br />
* [[List of United States high school mathematics competitions]]<br />
* [http://www.mathcounts.org/webarticles/anmviewer.asp?a=921&z=71 2006 MATHCOUNTS Countdown Video]</div>Dipenmhttps://artofproblemsolving.com/wiki/index.php?title=Asymptote_(Vector_Graphics_Language)&diff=57091Asymptote (Vector Graphics Language)2013-09-15T02:02:47Z<p>Dipenm: </p>
<hr />
<div>{{Asymptote}}<br />
<br />
'''Asymptote''' is a powerful vector graphics language designed for creating mathematical diagrams and figures. It can output images in either eps or pdf format, and is compatible with the standard mathematics typesetting language, [[LaTeX]]. It is also a complete programming language, and has cleaner syntax than its predecessor, [http://netlib.bell-labs.com/who/hobby/MetaPost.html MetaPost], which was a language used only for two-dimensional graphics.<br />
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Here is an example of an image that can be produced using Asymptote:<br />
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<center>[[Image:Figure1.jpg]]</center><br />
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In a sense, Asymptote is the ruler and compass of typesetting.<br />
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You can use Asymptote on the AoPSWiki right now, by enclosing the Asymptote code within <tt><nowiki><asy>...</asy></nowiki></tt> tags. For example, the following code<br />
<pre><nowiki><asy><br />
draw((0,0)--(3,7),red);<br />
dot((0,0));<br />
dot((3,7));<br />
label("Produced with Asymptote "+version.VERSION,point(S),2S);<br />
</asy></nowiki></pre><br />
created the picture <br />
<center><asy><br />
draw((0,0)--(3,7),red);<br />
dot((0,0));<br />
dot((3,7));<br />
label("Produced with Asymptote "+version.VERSION,point(S),2S);<br />
</asy></center><br />
And on the AoPS forums you can use <tt><nowiki>[asy]..[/asy]</nowiki></tt><br />
<pre><nowiki>[asy]<br />
draw((0,0)--(3,7),red);<br />
dot((0,0));<br />
dot((3,7));<br />
label("Produced with Asymptote "+version.VERSION,point(S),2S);<br />
[/asy]</nowiki></pre><br />
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[asy]<br />
pair A,B,C,X,Y,Z;<br />
A = (0,0);<br />
B = (1,0);<br />
C = (0.3,0.8);<br />
draw(A--B--C--A);<br />
X = (B+C)/2;<br />
Y = (A+C)/2;<br />
Z = (A+B)/2;<br />
draw(A--X, red);<br />
draw(B--Y,red);<br />
draw(C--Z,red);[/asy]<br />
=== See also ===<br />
*[[LaTeX]]<br />
* [http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php AoPS's Getting Started with LaTeX guide.]<br />
* <url>index.php?f=519 Asymptote Forum on AoPS</url><br />
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[[Asymptote: Getting Started | Next: Getting Started]]</div>Dipenm