https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Fuegocaliente&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T14:11:54ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=User:Fuegocaliente&diff=121842User:Fuegocaliente2020-04-30T01:43:27Z<p>Fuegocaliente: </p>
<hr />
<div>An AoPS user.<br />
Yup.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=121560User:Piphi2020-04-23T05:11:44Z<p>Fuegocaliente: /* User Count */</p>
<hr />
<div><center>[[File:Piphi-Avatar.png]]</center><br />
<br />
<div style="border:2px solid black; background:#eeeeee;"><br />
::::<font style="font-family: Verdana, sans-serif">[[User:Piphi|Userpage]] | [[User talk:Piphi|Talk]] | [[Special:Contributions/Piphi|Contributions]]</font><br />
</div><br />
<div style="border:2px solid black; background:#dddddd; align:center"><br />
==<font color="black" style="font-family: MV Boli, Verdana">User Count</font>==<br />
<font color="black">If this is your first time visting this page, edit it by incrementing the user count below by one.<br />
<br />
<center><font size="100px">19</font></center><br />
</font> <br />
</div><br />
<div style="border:2px solid black; background:#cccccc; align:center"><br />
<br />
==<font color="black" style="font-family: MV Boli, Verdana">About Me</font>==<br />
<font color="black">PM me if you want to find out about some cool things you can do with the AoPS wiki.<br />
<br />
My main project on the AoPS wiki is [[AoPS_Administrators#List_of_Admins | a list of all the AoPS admins]], everyone is welcome to add more admins to the list by clicking [https://artofproblemsolving.com/wiki/index.php?title=AoPS_Administrators&action=edit&section=1 here]. Soon you should also see me working on the [[Reaper Archives]].</font> <br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb; align:center"><br />
<br />
==<font color="black" style="font-family: MV Boli, Verdana">Contributions</font>==<br />
<br />
{| class="wikitable" style="background:#bbbbbb;"<br />
|+ <font color="black">Pages</font><br />
|- <!-- Start of a new row --><br />
| [[Administrators]] || [[2005 AMC 8 Problems/Problem 20]] || [[1997 IMO]]<br />
|-<br />
| [[AoPS Administrators]] || [[2007 iTest Problems/Problem 59]] || [[1998 IMO]]<br />
|-<br />
| [[AoPS Moderators]] || [[2007 iTest Problems/Problem 60]] || [[1999 IMO]]<br />
|-<br />
| [[Art of Problem Solving Academy]] || [[2007 iTest Problems/Problem TB1]] || [[2000 IMO]]<br />
|-<br />
| [[BBCode]] || [[2007 iTest Problems/Problem TB2]] || [[2001 IMO]]<br />
|-<br />
| [[BBCode:Tutorial]] || [[2007 iTest Problems/Problem TB3]] || [[2002 IMO]]<br />
|-<br />
| [[For The Win]] || [[2007 iTest Problems/Problem TB4]] || [[2003 IMO]]<br />
|-<br />
| [[Gmaas]] || [[2008 iTest Problems/Problem 95]] || [[2004 IMO]]<br />
|-<br />
| [[GMAAS]] || [[2015 IMO Problems/Problem 3]] || [[2005 IMO]]<br />
|-<br />
| [[IMO Shortlist Problems]] || [[2015 IMO Problems/Problem 5]] || [[2006 IMO]]<br />
|-<br />
| [[IMO Problems and Solutions]] || [[2015 IMO Problems/Problem 6]] || [[2007 IMO]]<br />
|-<br />
| [[IMO problems statistics since 2000]] || [[2016 IMO Problems/Problem 1]] || [[2009 IMO]]<br />
|-<br />
| [[IMO problems statistics]] || [[2016 IMO Problems/Problem 2]] || [[2010 IMO]]<br />
|-<br />
| [[International Mathematical Olympiad]] || [[2016 IMO Problems/Problem 4]] || [[2011 IMO]]<br />
|-<br />
| [[List of BBCodes]] || [[2016 IMO Problems/Problem 5]] || [[2012 IMO]]<br />
|-<br />
| [[Reflex angle]] || [[2016 IMO Problems/Problem 6]] || [[2015 IMO]]<br />
|-<br />
| [[Princeton University]] || [[2018 AIME I Problems]] || [[2016 IMO]]<br />
|-<br />
| [[Vieta's Formulas]] || [[2018 AMC 10A Problems/Problem 1]] || [[2017 IMO]]<br />
|-<br />
| [[2015 IMO Problems]] || [[2019 AMC 8 Problems/Problem 14]] || [[2018 IMO]]<br />
|-<br />
| [[2016 IMO Problems]] || [[2019 AMC 8 Problems/Problem 22]] || <br />
|}</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Fuegocaliente&diff=109663User talk:Fuegocaliente2019-09-03T18:48:15Z<p>Fuegocaliente: </p>
<hr />
<div>FuegoCaliente is an AoPS user. The username means hot fire in Spanish. The user decided that fire was hot, not cold, hence the username.<br />
<br />
You can call me fuego, fuego, or fuego (or fuegocaliente). If you were a counselor with me this summer then fugo works as well.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Fuegocaliente&diff=109662User talk:Fuegocaliente2019-09-03T18:47:57Z<p>Fuegocaliente: </p>
<hr />
<div>FuegoCaliente is an AoPS user. The username means hot fire in Spanish. The user decided that fire was hot, not cold, hence the username.<br />
<br />
You can call the user fuego, fuego, or fuego (or fuegocaliente). If you were a counselor with me this summer then fugo works as well.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Fuegocaliente&diff=109661User talk:Fuegocaliente2019-09-03T18:46:09Z<p>Fuegocaliente: /* ASDF */</p>
<hr />
<div>FuegoCaliente is an AoPS user. The username means hot fire in Spanish. The user decided that fire was hot, not cold, hence the username.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Fuegocaliente&diff=109660User talk:Fuegocaliente2019-09-03T18:45:41Z<p>Fuegocaliente: /* ASDF */ new section</p>
<hr />
<div>FuegoCaliente is an AoPS user. The username means hot fire in Spanish. The user decided that fire was hot, not cold, hence the username.<br />
<br />
== ASDF ==<br />
<br />
test</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2014_UMO_Problems&diff=982292014 UMO Problems2018-10-22T19:11:54Z<p>Fuegocaliente: /* Problem 5 */</p>
<hr />
<div>==Problem 1==<br />
<br />
Todd and Allison are playing a game on the grid shown below. At the beginning, an orange stone<br />
is placed in the center intersection on the grid. They take turns, with Todd going first. In each of<br />
Todd’s turns, he must move the orange stone from its current position to a horizontally or vertically<br />
adjacent intersection that is not occupied by a blue stone, and then he places a blue stone in the orange<br />
stone’s previous spot. In each of Allison’s turns, she places a blue stone on exactly one unoccupied<br />
intersection. Todd loses the game when he is forced to move into one of the corner intersections,<br />
labeled by <math>A, B, C</math>, and <math>D</math> in the diagram below. Allison loses if Todd can’t move.<br />
Allison tries to force Todd to lose in as few as turns as possible, and Todd tries to survive as long as<br />
possible. If both of them play as best they can, how many blue stones will be on the board at the end<br />
of the game? (You may assume that Todd always loses.)<br />
<br />
<asy><br />
size(200);<br />
for(int k=0;k<9;++k)<br />
{<br />
D((0,k)--(8,k),black);D((k,0)--(k,8),black);<br />
}<br />
MP("A",(0,8),NW);MP("B",(0,0),SW);MP("C",(8,0),SE);MP("D",(8,8),NE);<br />
D(CR((4,4),.2),black);<br />
fill(CR((4,4),.2),orange);<br />
</asy><br />
<br />
<br />
<br />
<br />
[[2014 UMO Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<br />
(a) Find all positive integers <math>x</math> and <math>y</math> that satisfy <cmath>x^2+y^2 = 2014,</cmath> or prove that there are no solutions.<br />
<br />
<br />
(b) Find all positive integers <math>x</math> and <math>y</math> that satisfy <cmath> x^2 + y^2 = 3222014,</cmath> or prove that there are no<br />
solutions.<br />
<br />
[[2014 UMO Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
Completely describe the set of all right triangles with positive integer-valued legs such that when four<br />
copies of the triangle are arranged in square formation shown below, the incenters of the four triangles<br />
lie on the extensions of the sides of the smaller square. (Note: the incenter of a triangle is the center<br />
of the circle inscribed in that triangle.)<br />
<br />
<asy><br />
size(200);<br />
path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));<br />
D(T,black);<br />
D(shift(10,0)*rotate(53)*T,black);<br />
D(shift(15,5)*rotate(233)*T,black);<br />
D(shift(15,0)*rotate(143)*T,black);<br />
D(shift(10,5)*rotate(323)*T,black);<br />
</asy><br />
<br />
[[2014 UMO Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
Joel is playing with ordered lists of integers in the following way. He starts out with an ordered list<br />
of nonnegative integers. Then, he counts the number of <math>0</math>’s, <math>1</math>’s, <math>2</math>’s, and so on in the list, writing<br />
the counts out as a new list. He stops counting when he has counted everything in the previous list.<br />
Then he takes the second list and applies the same process to get a third list. He repeats this process<br />
indefinitely.<br />
<br />
For example, he could start out with the ordered list <math>(0, 0, 0, 2)</math>. He counts three <math>0</math>’s, zero <math>1</math>’s, and one<br />
<math>2</math>, and then stops counting, so the second list is <math>(3, 0, 1)</math> In the second list, he counts one <math>0</math>, one <math>1</math>,<br />
zero <math>2</math>’s, and one <math>3</math>, so the third list is <math>(1, 1, 0, 1)</math>. Then he counts one <math>0</math> and three <math>1</math>’s, so the fourth list<br />
is <math>(1, 3)</math>. Here are the first few lists he writes down:<br />
<math>(0,0, 0, 2) \longrightarrow (3, 0,1) \longrightarrow (1, 1, 0, 1) \longrightarrow (1, 3) \longrightarrow \cdots </math><br />
If instead he started with <math>(0, 0)</math>, he would write down:<br />
<math>(0, 0) \longrightarrow (2) \longrightarrow (0,0, 1) \longrightarrow (2, 1) \longrightarrow \cdots</math><br />
If Joel starts out with an arbitrary list of nonnegative integers and then continues this process, there<br />
are certain lists <math>(m, n)</math> of length two that he might end up writing an infinite number of times. Find<br />
all such pairs <math>(m, n)</math>.<br />
<br />
[[2014 UMO Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Find all positive real numbers <math>x, y</math>, and <math>z</math> that satisfy both of the following equations.<br />
<cmath>\begin{align*} xyz & = 1\\<br />
x^2 + y^2 + z^2 & = 4x\sqrt{yz}- 2yz \end{align*}</cmath><br />
<br />
[[2014 UMO Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Draw <math>n</math> rows of <math>2n</math> equilateral triangles each, stacked on top of each other in a diamond shape, as<br />
shown below when <math>n = 3</math>. Set point <math>A</math> as the southwest corner and point <math>B</math> as the northeast corner.<br />
A step consists of moving from one point to an adjacent point along a drawn line segment, in one of<br />
the four legal directions indicated. A path is a series of steps, starting at <math>A</math> and ending at <math>B</math>, such<br />
that no line segment is used twice. One path is drawn below. Prove that for every positive integer <math>n</math>,<br />
the number of distinct paths is a perfect square. (Note: A perfect square is a number of the form <math>k^2</math>,<br />
where <math>k</math> is an integer).<br />
<br />
<asy><br />
size(200);<br />
path p1=polygon(3),p2=shift(sqrt(3)/2,1/2)*rotate(180)*polygon(3);<br />
for(int k=0;k<3;++k)<br />
{for (int j=0;j<3;++j)<br />
{draw(shift(sqrt(3)*(k+j/2),1.5*j)*p1,black+linewidth(.4));<br />
draw(shift(sqrt(3)*(k+j/2),1.5*j)*p2,black+linewidth(.4));}<br />
}<br />
pair O=(-sqrt(3)/2,-1/2),E=(sqrt(3),0),NW=(-sqrt(3)/2,3/2),NE=(sqrt(3)/2,3/2),SE=(sqrt(3)/2,-3/2);<br />
D(O--(O+E)--(O+E+NE)--(O+2E+NE)--(O+2E+NE+SE)--(O+2E+2NE+SE)--(O+2E+2NE+SE+NW)--(O+3E+2NE+SE+NW)--(O+3E+3NE+SE+NW),black+linewidth(2));<br />
MP("A",O,SW);<br />
MP("B",(O+3E+3NE+SE+NW),NE);<br />
draw(shift(10,0)*p1,black+linewidth(.4));<br />
draw(shift(10,sqrt(3))*p1,black+linewidth(.4));<br />
draw(shift(12.5,0)*p1,black+linewidth(.4));<br />
draw(shift(12.5,sqrt(3))*p1,black+linewidth(.4));<br />
MP("Legal Steps",(11.4,4.2));<br />
draw((10-sqrt(3)/2,-1/2)--(10,1.5-1/2),arrow=Arrow());<br />
draw((12.5-sqrt(3)/2,-1/2)--(12.5+sqrt(3)/2,-1/2),arrow=Arrow());<br />
draw((10,sqrt(3)+1)--(10+sqrt(3)/2,sqrt(3)-1/2),arrow=Arrow());<br />
draw((12.5+sqrt(3)/2,sqrt(3)-1/2)--(12.5,sqrt(3)+1),arrow=Arrow());<br />
</asy><br />
<br />
[[2014 UMO Problems/Problem 6|Solution]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=92478AMC historical results2018-02-28T23:56:28Z<p>Fuegocaliente: /* AMC 10A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
==2018==<br />
===AMC 10A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*DHR: TBD<br />
<br />
===AMC 10B===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*DHR: TBD<br />
<br />
===AMC 12A===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*DHR: TBD<br />
<br />
===AMC 12B===<br />
*Average score: TBD<br />
*AIME floor: TBD<br />
*DHR: TBD<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.56<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score:8<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 111<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 111<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 93<br />
*DHR: 111<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100.5<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100.5<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100.5<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
===AIME II===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: -100000000000000000000000000000 :P<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Moderators&diff=91302AoPS Moderators2018-02-16T18:15:32Z<p>Fuegocaliente: /* Moderators vs. Administrators */</p>
<hr />
<div>A moderator on the Art of Problem Solving website is a non-administrator user who has powers for a particular forum which non-moderators don't. That is: they can lock topics in a forum, move them, delete them, comment about certain posts and so on. The short form for moderator is mod, the short term for administrator is admin.<br />
<br />
When a user is an Administrator, their name shows up in red.<br />
When they are a mod, their name shows up in green.<br />
<br />
An user automatically becomes an administrator of a forum if they create it. This user can manage who gets to be a mod or admin in said forum. <br />
<br />
==Moderators vs. Administrators==<br />
<br />
Administrators are more powerful in Art of Problem Solving than moderators. Moderators answer to Administrators, and not vice versa.<br />
<br />
Some admins might have their names in blue instead of red. Those are AoPS site administrators. They are administrators to all forums on AoPS, even private forums and class forums. They are part of the AoPS ninja squad.<br />
<br />
==List of Admins ==<br />
<br />
Below is a list of some of the current AoPS site admins:<br />
<br />
<br />
5space (Dustin Ransom)<br />
ahuhn (Anika Huhn)<br />
ak.santana (Anakaren Santana)<br />
amysz (Amy Szczepanski)<br />
copeland (Jeremy Copeland)<br />
corinne (Corinne Madsen)<br />
devenware (Deven Ware)<br />
DuncanG (Duncan Gilles)<br />
douchman (Dorothy Couchman)<br />
gmaas (Grayson Maas)<br />
jbatterson (Jason Batterson)<br />
jgf1123 (James Fung)<br />
LauraZed (Laura Zehender)<br />
levans (Larry Evans)<br />
melschulz<br />
paulhryu (Paul Ryu)<br />
phxu (Phyllis Xu)<br />
psalerno (Paul Salerno)<br />
rrusczyk (Richard Rusczyk)<br />
rjashmore<br />
srogers (Shannon Rogers)<br />
sseraj (Samer Seraj)<br />
yatingliu (Yating Liu)</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Moderators&diff=91301AoPS Moderators2018-02-16T18:15:04Z<p>Fuegocaliente: /* List of Admins */</p>
<hr />
<div>A moderator on the Art of Problem Solving website is a non-administrator user who has powers for a particular forum which non-moderators don't. That is: they can lock topics in a forum, move them, delete them, comment about certain posts and so on. The short form for moderator is mod, the short term for administrator is admin.<br />
<br />
When a user is an Administrator, their name shows up in red.<br />
When they are a mod, their name shows up in green.<br />
<br />
An user automatically becomes an administrator of a forum if they create it. This user can manage who gets to be a mod or admin in said forum. <br />
<br />
==Moderators vs. Administrators==<br />
<br />
Administrators are more powerful in Art of Problem Solving than moderators. Moderators answer to Administrators, and not vice versa.<br />
<br />
Some admins might have their names in blue instead of red. Those are AoPS site administrators. They are administrators to all forums on AoPS, even private forums and class forums.<br />
<br />
==List of Admins ==<br />
<br />
Below is a list of some of the current AoPS site admins:<br />
<br />
<br />
5space (Dustin Ransom)<br />
ahuhn (Anika Huhn)<br />
ak.santana (Anakaren Santana)<br />
amysz (Amy Szczepanski)<br />
copeland (Jeremy Copeland)<br />
corinne (Corinne Madsen)<br />
devenware (Deven Ware)<br />
DuncanG (Duncan Gilles)<br />
douchman (Dorothy Couchman)<br />
gmaas (Grayson Maas)<br />
jbatterson (Jason Batterson)<br />
jgf1123 (James Fung)<br />
LauraZed (Laura Zehender)<br />
levans (Larry Evans)<br />
melschulz<br />
paulhryu (Paul Ryu)<br />
phxu (Phyllis Xu)<br />
psalerno (Paul Salerno)<br />
rrusczyk (Richard Rusczyk)<br />
rjashmore<br />
srogers (Shannon Rogers)<br />
sseraj (Samer Seraj)<br />
yatingliu (Yating Liu)</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User:Fuegocaliente&diff=91300User:Fuegocaliente2018-02-16T18:12:15Z<p>Fuegocaliente: </p>
<hr />
<div>An amazing AoPS user.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Fuegocaliente&diff=91299User talk:Fuegocaliente2018-02-16T18:10:43Z<p>Fuegocaliente: Created page with "FuegoCaliente is an AoPS user. The username means hot fire in Spanish. The user decided that fire was hot, not cold, hence the username."</p>
<hr />
<div>FuegoCaliente is an AoPS user. The username means hot fire in Spanish. The user decided that fire was hot, not cold, hence the username.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=AMC_10&diff=91298AMC 102018-02-16T18:08:29Z<p>Fuegocaliente: /* Resources */</p>
<hr />
<div>The '''American Mathematics Contest 10''' ('''AMC 10'''), along with the [[AMC 12]], is one of the first exams in the series of exams used to challenge bright students, grades 10 and below, on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO).<br />
<br />
High scoring AMC 10 and AMC 12 students are invited to take the [[American Invitational Mathematics Examination]] (AIME).<br />
<br />
The AMC 10 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC.<br />
<br />
<br />
== Format ==<br />
<br />
The AMC 10 is a 25 question, 75 minute multiple choice test. Problems generally increase in difficulty as the exam progresses. Calculators were permitted on old AMC tests; however, as of 2008 and later, calculators are not permitted for use.<br />
<br />
The AMC 10 is scored in a way that penalizes guesses. Correct answers are worth 6 points, incorrect questions are worth 0 points, and unanswered answers are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, unanswered questions were awarded 2.5 points. In 2006 and 2007, unanswered questions were awarded 2 points. Students that are in the top 2.5% of the AMC 10 contest are invited to take the [[AIME]].[http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html]<br />
<br />
== Curriculum ==<br />
The AMC 10 tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solved by students without any background in calculus or trigonometry.<br />
<br />
== Resources ==<br />
=== Links ===<br />
* [http://www.maa.org/math-competitions AMC homepage], their [http://www.unl.edu/amc/e-exams/e5-amc10/amc10.shtml AMC 10 page], and [http://www.unl.edu/amc/mathclub/index.html practice problems]<br />
* The [[AoPS]] [http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AMC10.php AMC 10 guide] (Warning: Scoring system is out of date) <br />
* [https://artofproblemsolving.com/community/c5_contests_amp_programs Contest and Programs Forum] for discussion of the AMC and problems from AMC exams, and any other competition.<br />
* [[AMC 10 Problems and Solutions | Past AMC 10 exams]].<br />
<br />
=== Recommended reading ===<br />
* [http://www.artofproblemsolving.com/Store/contests.php?contest=amc Problem and solution books for past AMC exams].<br />
<br />
* [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:algebra Introduction to Algebra] by [[Richard Rusczyk]]. <br />
* [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:counting Introduction to Counting & Probability] by Dr. [[David Patrick]]. <br />
* [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:geometry Introduction to Geometry] by [[Richard Rusczyk]]. <br />
* [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:nt Introduction to Number Theory] by [[Mathew Crawford]].<br />
* [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops1 The Art of Problem Solving Volume 1] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br />
* [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops2 The Art of Problem Solving Volume 2] by [[Richard Rusczyk]] and [[Sandor Lehoczky]].<br />
<br />
=== AMC Preparation Classes ===<br />
'''These should be taken if a student is having trouble or wants a more clear and confirmed understanding of concepts.'''<br />
* [[AoPS]] hosts an [http://www.artofproblemsolving.com/School/index.php online school] teaching introductory classes in topics covered by the AMC 10 as well as an AMC 10 preparation class.<br />
* [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AMC 10 and AMC 12. [http://www.artofproblemsolving.com/School/mathjams.php Math Jam Schedule]<br />
* [[EPGY]] offers an AMC 10 preparation class.<br />
<br />
== See also ==<br />
* [[Mathematics competitions]]<br />
* [[ARML]]<br />
* [[Mathematics summer programs]]<br />
<br />
<br />
<br />
[[Category:Mathematics competitions]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Moderators&diff=89109AoPS Moderators2017-12-23T20:13:15Z<p>Fuegocaliente: /* List of Admins */</p>
<hr />
<div>A moderator on the Art of Problem Solving website is a non-administrator user who has powers for a particular forum which non-moderators don't. That is: they can lock topics in a forum, move them, delete them, comment about certain posts and so on. The short form for moderator is mod, the short term for administrator is admin.<br />
<br />
When a user is an Administrator, their name shows up in red.<br />
When they are a mod, their name shows up in green.<br />
<br />
An user automatically becomes an administrator of a forum if they create it. This user can manage who gets to be a mod or admin in said forum. <br />
<br />
==Moderators vs. Administrators==<br />
<br />
Administrators are more powerful in Art of Problem Solving than moderators. Moderators answer to Administrators, and not vice versa.<br />
<br />
Some admins might have their names in blue instead of red. Those are AoPS site administrators. They are administrators to all forums on AoPS, even private forums and class forums.<br />
<br />
==List of Admins ==<br />
<br />
Below is a list of some of the current AoPS site admins:<br />
<br />
<br />
5space (Dustin Ransom)<br />
ahuhn (Anika Huhn)<br />
ak.santana (Anakaren Santana)<br />
amysz (Amy Szczepanski)<br />
corinne (Corinne Madsen)<br />
devenware (Deven Ware)<br />
DuncanG (Duncan Gilles)<br />
gmaas (Grayson Maas)<br />
jgf1123 (James Fung)<br />
LauraZed (Laura Zehender)<br />
levans (Larry Evans)<br />
paulhryu (Paul Ryu)<br />
phxu (Phyllis Xu)<br />
psalerno (Paul Salerno)<br />
srogers (Shannon Rogers)<br />
sseraj (Samer Seraj)<br />
copeland (Jeremy Copeland)</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=88309AMC historical results2017-11-17T22:42:10Z<p>Fuegocaliente: /* 2017 */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.55<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 56.99<br />
*AIME floor: 96<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: <br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: <br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 97.5<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 117<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100 (Top 5% 106.5)<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User:Slester&diff=87956User:Slester2017-10-24T22:45:59Z<p>Fuegocaliente: Created page with "Stephen joined AoPS in 2015. He graduated with a degree in mathematics from the University of Tennessee, Knoxville, with a minor in French, but 日本語も話せます。Sinc..."</p>
<hr />
<div>Stephen joined AoPS in 2015. He graduated with a degree in mathematics from the University of Tennessee, Knoxville, with a minor in French, but 日本語も話せます。Since then, he has taught English in France, programmed in French in Montreal, and has traveled from coast to coast within the United States. In his free time, he enjoys looking at pictures of sloths on the Internet, learning languages, playing video and board games, arcane sorcery, digging through dusty vinyl record stores, and installing and ultimately reinstalling Linux distributions.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User:Rrusczyk&diff=87955User:Rrusczyk2017-10-24T22:45:25Z<p>Fuegocaliente: Created page with "Art of Problem Solving was founded by Richard Rusczyk in 2003 to create interactive educational opportunities for avid math students. Richard is one of the co-authors of the A..."</p>
<hr />
<div>Art of Problem Solving was founded by Richard Rusczyk in 2003 to create interactive educational opportunities for avid math students. Richard is one of the co-authors of the Art of Problem Solving classic textbooks, author of Art of Problem Solving's Introduction to Algebra, Introduction to Geometry, and Precalculus textbooks, co-author of Art of Problem Solving's Intermediate Algebra and Prealgebra, one of the co-creators of the Mandelbrot Competition, and a past Director of the USA Mathematical Talent Search. He was a participant in National MATHCOUNTS, a three-time participant in the Math Olympiad Summer Program, and a USA Mathematical Olympiad winner (1989). He received the World Federation of National Mathematics Competitions Paul Erdös Award in 2014. He graduated from Princeton University in 1993, and worked as a bond trader for D.E. Shaw & Company for four years. AoPS marks Richard's return to his vocation - educating motivated students.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Ellipse&diff=87485Ellipse2017-09-13T02:07:02Z<p>Fuegocaliente: /* Related Formulae */</p>
<hr />
<div>An '''ellipse''' is a type of [[conic section]].<br />
<br />
==Definition==<br />
An ellipse is formed by cutting through a [[cone]] at an [[angle]]. <br />
Equivalently, it is defined as the [[locus]], or [[set]], of all [[point]]s <math>P</math> such that the sum of the distances from <math>P</math> to two fixed [[focus|foci]] is a constant. (The equivalence of these two definitions is a non-trivial fact.)<br />
<br />
==Intuition==<br />
Ellipses tend to resemble [[circle]]s which have been "flattened" or "stretched." They occur in nature as well as in mathematics: as was proven in [[Kepler's Laws]], the planets all revolve about the sun in elliptical, not circular, orbits with the sun at one of the foci. Note that the circle is just a special case of the ellipse, just as a square is to a rectangle, and occurs when (in the first definition) the cutting plane is [[perpendicular]] to the axis of the the cone, or (in the second definition) the two foci of the ellipse coincide.<br />
<br />
Using the second definition of an ellipse given above, one may easily construct an ellipse from household materials. To draw an ellipse with two pushpins, a loop of string, pencil, and paper, stick the pushpins in the paper place the string on the paper so that both pushpins are inside it. The pushpins will be the foci of the ellipse, and the length of the string will determine the sum of the distances from a point on the ellipse to the two foci. Hold the pencil on the paper such that the string is taut against the pencil tip and the two pushpins. Then move the pencil tip while keeping the string taut. This will trace out an ellipse.<br />
<br />
==Related Terminology==<br />
For a given non-circular ellipse, there will be two points on the ellipse closest to the center and two points furthest away -- it will be "tall and skinny" or "short and fat." The segment connecting the center of the ellipse to one of the "farther away ends" is called the ''[[semimajor axis]]'' and the segment connecting the center to a closer end is called the ''[[semiminor axis]]''. These two segments are perpendicular. Drawing all four semi-axes divides the ellipse into 4 [[congruent (geometry)|congruent]] quarters.<br />
<br />
{{asy image|<br />
<asy><br />
size(200);<br />
D((-5,0)--(0,0)--(0,-3));<br />
MC(90,"\mbox{semiminor axis}",7,D((0,0)--(0,3),green+linewidth(1)),E);<br />
MC("\mbox{semimajor axis}",7,D((0,0)--(5,0),red+linewidth(1)),S);<br />
D(ellipse(D("\mbox{center}",7,(0,0),black,SW),5,3),orange+linewidth(1));<br />
</asy>|right|Ellipse<br />
}}<br />
<br />
An ellipse in a [[Cartesian coordinate system]] with center <math>C = (h, k)</math> whose axes are parallel to the coordinate axes, with the horizontal semi-axis of length <math>a</math> and the vertical semi-axis of length <math>b</math> is given by the equation <math>\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1</math>. In particular, if the center of the ellipse is the origin this simplifies to <math>\frac{x^2}{a^2}+\frac{y^2}{b^2}=1</math>.<br />
<br />
The three-dimensional counterpart of the ellipse is the [[ellipsoid]].<br />
<br />
==Properties==<br />
* Let <math>F_{1},F_{2}</math> be the foci of an ellipse, let <math>P</math> be a point on the ellipse, and let <math>\ell</math> be the tangent line to the ellipse at <math>P</math>. Then it is true that the bisector of angle <math>F_{1}PF_{2}</math> is perpendicular to <math>\ell</math>.<br />
<asy><br />
size(200);<br />
defaultpen(fontsize(8));<br />
pair P=(3,12/5), F1=(-4,0), F2=(4,0);<br />
D(ellipse((0,0),5,3));<br />
D((-1,21/5)--(7,3/5));<br />
D(F1--P--F2);<br />
D(P--P-(9/5,4));<br />
dot(P^^F1^^F2);<br />
label("$P$",P,(1,1));label("$F_{1}$",F1,(0,-2));label("$F_{2}$",F2,(0,-2));label("$\ell$",(7,3/5),(0,2));<br />
</asy><br />
<br />
==Related Formulae==<br />
*The [[area]] of an ellipse with semimajor and semiminor axes <math>a,b</math> is <math>ab\pi</math>.<br />
*The [[circumference]] of an ellipse is <math>4aE(\epsilon)</math>, where the <math>E</math> is the second [[elliptic integral]].<br />
<br />
==Problems==<br />
===Introductory===<br />
<br />
*The ellipse with axis lengths <math>14</math> and <math>16</math> has the general equation of <math>\frac{x^2}{a^2}+\frac{y^2}{b^2}=1</math>. Find the value of <math>a^2+b^2</math>.<br />
<br />
===Intermediate===<br />
*Let <math> w_1 </math> and <math> w_2 </math> denote the circles <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest possible value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally tangent to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math> ([[2005 AIME II Problems/Problem 15|Source]])<br />
*An equilateral triangle is inscribed in the ellipse whose equation is <math>x^2+4y^2=4</math>. One vertex of the triangle is <math>(0,1)</math>, one altitude is contained in the y-axis, and the length of each side is <math>\sqrt{\frac mn}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. ([[2001 AIME I Problems/Problem 5|Source]])<br />
*An [[ellipse]] has [[focus | foci]] at <math>(9,20)</math> and <math>(49,55)</math> in the <math>xy</math>-plane and is [[tangent line | tangent]] to the <math>x</math>-axis. What is the length of its [[major axis]]? ([[1985 AIME Problems/Problem 11|Source]])<br />
===Olympiad===<br />
<br />
==See also==<br />
* [[Parabola]]<br />
* [[Conic section]]s<br />
* [[Geometry]]<br />
* [[Polynomial]]s<br />
<br />
[[Category:Definition]]<br />
[[CAtegory:Geometry]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Proofs&diff=85676Proofs2017-05-15T18:16:07Z<p>Fuegocaliente: /* Pythagorean Theorem */</p>
<hr />
<div>==Quadratic Formula==<br />
<br />
Let <math>ax^2+bx+c=0</math>. Then<br />
<cmath>x^2+\frac{b}{a}x+\frac{c}{a}=0</cmath><br />
Completing the square, we get<br />
<cmath>\left(x+\frac{b}{2a}\right)^2 +~ \frac{b^2-4ac}{4a^2}=0 \Rightarrow x~+~\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm \sqrt{b^2-4ac}}{2a}</cmath><br />
Simplifying, we see<br />
<cmath>\boxed{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}</cmath><br />
<br />
==Pythagorean Theorem==<br />
<br />
http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif<br />
<br />
Since area of green square is <math>a^2</math><br />
<br />
Since are of blue square is <math>b^2</math><br />
<br />
Since red square is <math>c^2</math><br />
<br />
We have the following relationship<br />
<br />
Based on this, we get we get <math>\boxed {a^2+b^2=c^2}</math> (here we get the Pythagorean Theorem)</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Proofs&diff=85675Proofs2017-05-15T18:15:58Z<p>Fuegocaliente: /* Pythagorean Theorem */</p>
<hr />
<div>==Quadratic Formula==<br />
<br />
Let <math>ax^2+bx+c=0</math>. Then<br />
<cmath>x^2+\frac{b}{a}x+\frac{c}{a}=0</cmath><br />
Completing the square, we get<br />
<cmath>\left(x+\frac{b}{2a}\right)^2 +~ \frac{b^2-4ac}{4a^2}=0 \Rightarrow x~+~\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm \sqrt{b^2-4ac}}{2a}</cmath><br />
Simplifying, we see<br />
<cmath>\boxed{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}</cmath><br />
<br />
==Pythagorean Theorem==<br />
<br />
[img]http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif[/img]<br />
<br />
Since area of green square is <math>a^2</math><br />
<br />
Since are of blue square is <math>b^2</math><br />
<br />
Since red square is <math>c^2</math><br />
<br />
We have the following relationship<br />
<br />
Based on this, we get we get <math>\boxed {a^2+b^2=c^2}</math> (here we get the Pythagorean Theorem)</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:Tutorial&diff=84707AoPS Wiki:Tutorial2017-03-15T17:58:52Z<p>Fuegocaliente: /* HTML */</p>
<hr />
<div>{{shortcut|[[A:T]]}}<br />
This '''AoPSWiki tutorial''' will help guide you through the process of learning to contribute to the [[AoPSWiki]]. Also see the categories [[:Category:AoPSWiki | AoPSWiki]] and [[:Category:Tutorials | Tutorials]] for more information on certain topics.<br />
<br />
In addition to the information below, there is a good [http://en.wikipedia.org/wiki/Help:Editing Wikipedia editing page] which contains most of the information below and will be incorporated into this tutorial over time.<br />
<br />
<br />
== Editing the AoPSWiki ==<br />
The AoPSWiki is meant to be edited by you, the users. If you ever find yourself saying that you could have done a better job on an article, we encourage you to do so! Before you can edit the article, however, you should make sure you are familiar with how the AoPSWiki works.<br />
<br />
=== Learning About AoPSWiki ===<br />
All users are encouraged to read up on what a wiki is and look around AoPSWiki. Click the [[Special:Random | Random Page]] button on the navigation panel at right a few times to get a feel for the structure of content in the AoPSWiki. Try clicking through a few of the colored links on some of the pages. Some will take you to other parts of the AoPSWiki and some will take you out of the AoPSWiki to other websites.<br />
<br />
Read about [[AoPSWiki:What makes AoPSWiki different|what makes AoPSWiki different]].<br />
<br />
=== Creating an Article ===<br />
If you search a topic or click on a link to an article that is not yet written, you yourself can begin creating that article. It is preferred to make bold the name of the article when it first appears in the article, particularly for terms that are given definitions. Bolding is explained in the Wiki Syntax section below. Read [http://artofproblemsolving.com/community/c65h98050p553863 Tips for Article Creation] to better understand the AoPSWiki philosophy of article content.<br />
<br />
If you can't think of a good topic to write about, look through the list of [[:Category:stubs|stubs]] for articles which already exist but have very little content.<br />
<br />
=== Editing an Article ===<br />
Click on the '''Edit''' button above an article while you are logged into the AoPSWiki. If the article is new, you can just start writing. If the article already exists, you will see the article text, [[MediaWiki]] syntax included. Write around or over the text to change the content of the article.<br />
<br />
If you would like to experiment with the Wiki format, please use the [http://en.wikipedia.org/wiki/Wikipedia:Sandbox Sandbox] to play around with new features.<br />
<br />
=== Wiki Syntax ===<br />
First, while in edit mode, there is a control panel in the upper left of the text box. Hover your cursor over the buttons to get an idea as to what they do. Remember that you can test syntax on your own user page. You can also hunt for examples by going into edit mode while in any article. Next, you can practice using wiki syntax in the [[sandbox]]. Here are a few common ways to format content in your article:<br />
* To create '''bullets''' link for this list, use the asterisk (*) at the beginning of each line.<br />
* Use the colon (:) to '''indent'''.<br />
* To ''italicize'' text, place two apostrophes both before and after the <nowiki>''text''</nowiki>.<br />
* To '''bold''' text, place three apostrophes both before and after the <nowiki>'''text'''</nowiki>.<br />
<br />
==== Internal Links ====<br />
Internal links are at the heart of what makes a wiki. Place two square brackets on each side of a word or phrase to link the text to the article of the same name. In order to create the link [[mathematics]], you would type <nowiki>[[mathematics]]</nowiki>. Also, for the cases when it is necessary to make the link-forming part of the phrase different from the actual title of the article it refers to, you may use the piped link format like in <nowiki>[[real number|reals]]</nowiki>, which will create a link to the article titled '''Real number''' but show and highlight the word '''reals''' instead. If you wish to refer to a specific section of an article, say that Natural Logarithm section of the Logarithm article, you use <nowiki>[[Logarithm#Natural Logarithm|natural logarithm]]</nowiki>, which comes out like this: [[Logarithm#Natural Logarithm|natural logarithm]].<br />
<br />
==== External Links ====<br />
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[[Category:AoPSWiki|Tutorial]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=User:Fuegocaliente&diff=84314User:Fuegocaliente2017-03-01T23:38:46Z<p>Fuegocaliente: Created page with "An amazing user"</p>
<hr />
<div>An amazing user</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_6&diff=829732017 AMC 10A Problems/Problem 62017-02-08T22:14:39Z<p>Fuegocaliente: /* See Also */</p>
<hr />
<div></div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_6&diff=829722017 AMC 10A Problems/Problem 62017-02-08T22:14:31Z<p>Fuegocaliente: /* Solution */</p>
<hr />
<div>==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_24&diff=829002017 AMC 10A Problems/Problem 242017-02-08T21:35:13Z<p>Fuegocaliente: /* Solution */</p>
<hr />
<div>==Problem==<br />
For certain real numbers <math>a</math>, <math>b</math>, and <math>c</math>, the polynomial <cmath>g(x) = x^3 + ax^2 + x + 10</cmath>has three distinct roots, and each root of <math>g(x)</math> is also a root of the polynomial <cmath>f(x) = x^4 + x^3 + bx^2 + 100x + c.</cmath>What is <math>f(1)</math>?<br />
<br />
<math>\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005</math><br />
<br />
==Solution==<br />
<math>f(x)</math> must have four roots, three of which are roots of <math>g(x)</math>. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of <math>f(x)</math> and <math>g(x)</math> are the same, we know that<br />
<br />
<cmath>f(x)=g(x)(x-r)</cmath><br />
<br />
where <math>r\in\mathbb{C}</math> is the fourth root of <math>f(x)</math>. Substituting <math>g(x)</math> and expanding, we find that<br />
<br />
<cmath>\begin{align*}f(x)&=(x^3+ax^2+x+10)(x-r)\\<br />
&=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.\end{align*}</cmath><br />
<br />
Comparing coefficients with <math>f(x)</math>, we see that<br />
<br />
<cmath>\begin{align*}<br />
a-r&=1\\<br />
1-ar&=b\\<br />
10-r&=100\\<br />
-10r&=c\\<br />
\end{align*}</cmath><br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_6&diff=828992017 AMC 10A Problems/Problem 62017-02-08T21:34:16Z<p>Fuegocaliente: /* Problem */</p>
<hr />
<div>==Solution==<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_10&diff=827872017 AMC 10A Problems/Problem 102017-02-08T19:16:46Z<p>Fuegocaliente: Created page with "==Problem== Joy has <math>30</math> thin rods, one each of every integer length from <math>1</math> cm through <math>30</math> cm. She places the rods with lengths <math>3</ma..."</p>
<hr />
<div>==Problem==<br />
Joy has <math>30</math> thin rods, one each of every integer length from <math>1</math> cm through <math>30</math> cm. She places the rods with lengths <math>3</math> cm, <math>7</math> cm, and <math>15</math> cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?<br />
<br />
<math>\text{(A) 16}\qquad\text{(B) 17}\qquad\text{(C) 18}\qquad\text{(D) 19}\qquad\text{(E) 20}</math><br />
<br />
==Solution==<br />
The triangle inequality generalizes to all polygons, so you can just use <math>x < 3+7+15</math> and <math>x+3+7>15</math> to get <math>5<x<25</math>, so <math>x</math> takes on 19 values. Then, of course, you subtract 2 because 7 and 15 are already used, to get <math>\boxed{17}</math>.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=827862017 AMC 10A Problems/Problem 42017-02-08T19:14:22Z<p>Fuegocaliente: /* Problem */</p>
<hr />
<div>==Problem==<br />
Mia is “helping” her mom pick up <math>30</math> toys that are strewn on the floor. Mia’s mom manages to put <math>3</math> toys into the toy box every <math>30</math> seconds, but each time immediately after those <math>30</math> seconds have elapsed, Mia takes <math>2</math> toys out of the box. How much time, in minutes, will it take Mia and her mom to put all <math>30</math> toys into the box for the first time?<br />
<math>\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5</math><br />
<br />
==Solution==<br />
Every <math>30</math> seconds <math>3-2</math> toys are put in the box, so after <math>27\cdot30seconds</math> there will be <math>27</math> toys in the box. Mia's mom will then put <math>3</math> toys into to the box and we have our total amount of time to be <math>27\cdot30+30=840</math> seconds, which equals <math>14</math> minutes. <math>\boxed{(B) 14}</math></div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=827782017 AMC 10A Problems/Problem 142017-02-08T19:04:47Z<p>Fuegocaliente: Created page with "==Problem== Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was <math>A</math> dollars. The cost of his movie ticket was..."</p>
<hr />
<div>==Problem==<br />
Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was <math>A</math> dollars. The cost of his movie ticket was <math>20\%</math> of the difference between <math>A</math> and the cost of his soda, while the cost of his soda was <math>5\%</math> of the difference between <math>A</math> and the cost of his movie ticket. To the nearest whole percent, what fraction of <math>A</math> did Roger pay for his movie ticket and soda?<br />
<br />
<math> \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%</math></div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_23&diff=827722017 AMC 10A Problems/Problem 232017-02-08T19:00:23Z<p>Fuegocaliente: Created page with "==Problem== How many triangles with positive area have all their vertices at points <math>(i,j)</math> in the coordinate plane, where <math>i</math> and <math>j</math> are int..."</p>
<hr />
<div>==Problem==<br />
How many triangles with positive area have all their vertices at points <math>(i,j)</math> in the coordinate plane, where <math>i</math> and <math>j</math> are integers between <math>1</math> and <math>5</math>, inclusive?<br />
<br />
<math>\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300</math></div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_24&diff=827692017 AMC 10A Problems/Problem 242017-02-08T18:59:27Z<p>Fuegocaliente: Created page with "==Problem== For certain real numbers <math>a</math>, <math>b</math>, and <math>c</math>, the polynomial <cmath>g(x) = x^3 + ax^2 + x + 10</cmath>has three distinct roots, and..."</p>
<hr />
<div>==Problem==<br />
For certain real numbers <math>a</math>, <math>b</math>, and <math>c</math>, the polynomial <cmath>g(x) = x^3 + ax^2 + x + 10</cmath>has three distinct roots, and each root of <math>g(x)</math> is also a root of the polynomial <cmath>f(x) = x^4 + x^3 + bx^2 + 100x + c.</cmath>What is <math>f(1)</math>?<br />
<br />
<math>\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005</math></div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=827672017 AMC 10A Problems/Problem 42017-02-08T18:55:54Z<p>Fuegocaliente: Created page with "==Problem== Mia is “helping” her mom pick up <math>30</math> toys that are strewn on the floor. Mia’s mom manages to put <math>3</math> toys into the toy box every <math..."</p>
<hr />
<div>==Problem==<br />
Mia is “helping” her mom pick up <math>30</math> toys that are strewn on the floor. Mia’s mom manages to put <math>3</math> toys into the toy box every <math>30</math> seconds, but each time immediately after those <math>30</math> seconds have elapsed, Mia takes <math>2</math> toys out of the box. How much time, in minutes, will it take Mia and her mom to put all <math>30</math> toys into the box for the first time?<br />
<math>\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5</math></div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Pythagorean_Theorem&diff=82454Pythagorean Theorem2017-01-20T20:34:24Z<p>Fuegocaliente: /* Common Pythagorean Triples */</p>
<hr />
<div>The '''Pythagorean Theorem''' states that for a [[right triangle]] with legs of length <math>a</math> and <math>b</math> and [[hypotenuse]] of length <math>c</math> we have the relationship <math>{a}^{2}+{b}^{2}={c}^{2}</math>. This theorem has been know since antiquity and is a classic to prove; hundreds of proofs have been published and many can be demonstrated entirely visually(the book ''The Pythagorean Proposition'' alone consists of more than 370). The Pythagorean Theorem is one of the most frequently used theorems in [[geometry]], and is one of the many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theorem. <br />
<br />
This is generalized by the [[Geometric inequality#Pythagorean_Inequality | Pythagorean Inequality]] and the [[Law of Cosines]].<br />
<br />
== Proofs ==<br />
<br />
In these proofs, we will let <math>ABC </math> be any right triangle with a right angle at <math>{} C </math>.<br />
<br />
=== Proof 1 ===<br />
<br />
We use <math>[ABC] </math> to denote the area of triangle <math>ABC </math>.<br />
<br />
Let <math>H </math> be the perpendicular to side <math>AB </math> from <math>{} C </math>.<br />
<br />
<center><br />
<asy><br />
pair A, B, C, H;<br />
A = (0, 0);<br />
B = (4, 3);<br />
C = (4, 0);<br />
H = foot(C, A, B);<br />
<br />
draw(A--B--C--cycle);<br />
draw(C--H);<br />
draw(rightanglemark(A, C, B));<br />
draw(rightanglemark(C, H, B));<br />
label("$A$", A, SSW);<br />
label("$B$", B, ENE);<br />
label("$C$", C, SE);<br />
label("$H$", H, NNW);<br />
</asy><br />
</center><br />
<br />
Since <math>ABC, CBH, ACH</math> are similar right triangles, and the areas of similar triangles are proportionate to the squares of corresponding side lengths,<br />
<center><br />
<math> \frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2} </math>.<br />
</center><br />
But since triangle <math>ABC </math> is composed of triangles <math>CBH </math> and <math>ACH </math>, <math>[ABC] = [CBH] + [ACH] </math>, so <math>AB^2 = CB^2 + AC^2 </math>. {{Halmos}}<br />
<br />
=== Proof 2 ===<br />
<br />
Consider a circle <math>\omega </math> with center <math>B </math> and radius <math>BC </math>. Since <math>BC </math> and <math>AC </math> are perpendicular, <math>AC </math> is tangent to <math>\omega </math>. Let the line <math>AB </math> meet <math>\omega </math> at <math>Y </math> and <math>X </math>, as shown in the diagram:<br />
<br />
<center>[[Image:Pyth2.png]]</center><br />
<br />
Evidently, <math>AY = AB - BC </math> and <math>AX = AB + BC </math>. By considering the [[power of a point | power]] of point <math>A </math> with respect to <math>\omega </math>, we see<br />
<br />
<center><br />
<math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>. {{Halmos}}<br />
</center><br />
<br />
== Common Pythagorean Triples ==<br />
A [[Pythagorean Triple]] is a [[set]] of 3 [[positive integer]]s such that <math>a^{2}+b^{2}=c^{2}</math>, i.e. the 3 numbers can be the lengths of the sides of a right triangle. Among these, the [[Primitive Pythagorean Triple]]s, those in which the three numbers have no common [[divisor]], are most interesting. A few of them are:<br />
<br />
<cmath>3-4-5</cmath><br />
<cmath>5-12-13</cmath><br />
<cmath>9-12-15</cmath><br />
<cmath>7-24-25</cmath><br />
<cmath>8-15-17</cmath><br />
<cmath>9-40-41</cmath><br />
<cmath>12-35-37</cmath><br />
<cmath>20-21-29</cmath><br />
<br />
== Problems ==<br />
=== Introductory ===<br />
* [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]<br />
* [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]]<br />
<br />
== External links ==<br />
*[http://www.cut-the-knot.org/pythagoras/index.shtml 75 proofs of the Pythagorean Theorem]<br />
<br />
[[Category:Geometry]]<br />
<br />
[[Category:Theorems]]<br />
<br />
[[Category:Geometry]]<br />
[[Category:Pythagorean Theorem]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_5&diff=819632016 AMC 8 Problems/Problem 52016-12-21T23:38:19Z<p>Fuegocaliente: /* Solution 1 */</p>
<hr />
<div>The number <math>N</math> is a two-digit number.<br />
<br />
• When <math>N</math> is divided by <math>9</math>, the remainder is <math>1</math>.<br />
<br />
• When <math>N</math> is divided by <math>10</math>, the remainder is <math>3</math>.<br />
<br />
What is the remainder when <math>N</math> is divided by <math>11</math>?<br />
<br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7</math><br />
<br />
==Solution 1==<br />
From the second bullet point, we know that the second digit must be <math>3</math>. Because there is a remainder of <math>1</math> when it is divided by <math>9</math>, the multiple of <math>9</math> must end in a <math>2</math>. We now look for this one: <br />
<br />
<math>9(1)=9\\<br />
9(2)=18\\<br />
9(3)=27\\<br />
9(4)=36\\<br />
9(5)=45\\<br />
9(6)=54\\<br />
9(7)=63\\<br />
9(8)=72</math><br />
<br />
The number <math>72+1=73</math> satisfies both conditions. We subtract the biggest multiple of <math>11</math> less than <math>73</math> to get the remainder. Thus, <math>73-11(6)=73-66=\boxed{\textbf{(E) }7}</math>.<br />
<br />
{{AMC8 box|year=2016|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_25&diff=819622010 AMC 10B Problems/Problem 252016-12-21T23:36:42Z<p>Fuegocaliente: /* See also */</p>
<hr />
<div>== Problem ==<br />
Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that<br />
<br />
<center><br />
<math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/><br />
<math>P(2) = P(4) = P(6) = P(8) = -a</math>.<br />
</center><br />
<br />
What is the smallest possible value of <math>a</math>?<br />
<br />
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math><br />
<br />
== Solution ==<br />
We observe that because <math>P(1) = P(3) = P(5) = P(7) = a</math>, if we define a new polynomial <math>R(x)</math> such that <math>R(x) = P(x) - a</math>, <math>R(x)</math> has roots when <math>P(x) = a</math>; namely, when <math>x=1,3,5,7</math>.<br />
<br />
Thus since <math>R(x)</math> has roots when <math>x=1,3,5,7</math>, we can factor the product <math>(x-1)(x-3)(x-5)(x-7)</math> out of <math>R(x)</math> to obtain a new polynomial <math>Q(x)</math> such that <math>(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a</math>.<br />
<br />
Then, plugging in values of <math>2,4,6,8,</math> we get <br />
<br />
<cmath>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</cmath><br />
<cmath>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a</cmath><br />
<cmath>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</cmath><br />
<cmath>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</cmath><br />
<br />
<math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</math><br />
Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>.<br />
Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.<br />
<br />
== Critique ==<br />
<br />
The above solution is incomplete. What is really proven is that 315 is a factor of <math>a</math>, if such an <math>a</math> exists. That only rules out answer A.<br />
<br />
To prove that the answer is correct, one could exhibit a polynomium that satisfies the requirements with <math>a=315</math>. Here's one: <math>P(x) = -8x^7 + 252 x^6 -3248x^5 + 22050x^4 -84392x^3 + 179928x^2-194592x+80325</math>.<br />
<br />
You get that from the <math>8\times8</math> matrix <math>M_{i,j} = i^{j-1}</math> and <math>y^T=(315,-315,\ldots,315,-315)</math> and computing <math>c=M^{-1}y</math> which comes out as the all-integer coefficients above.<br />
<br />
== Critique of Critique==<br />
<br />
First of all, the solution shows that <math>a</math> is a multiple of <math>315</math>, not a factor of <math>315</math>. Many people confuse the usage of the words 'factor' and 'multiple'. Secondly, even if <math>a</math> is a multiple of <math>315</math>, it does not mean that you can instantly get that the answer is <math>315</math> because we need to know that <math>a=315</math> is possible. After all, <math>a</math> is also a multiple of <math>1</math>, but <math>1</math> is definitely not the smallest possible number.<br />
<br />
To complete the solution, we can let <math>a = 315</math>, and then try to find <math>Q(x)</math>. We know from the above calculation that <math>Q(2)=42, Q(4)=-70, Q(6)=42</math>, and <math>Q(8)=-6</math>. Then we can let <math>Q(x) = T(x)(x-2)(x-6)+42</math>, getting <math>T(4)=28, T(8)=-4</math>. Let <math>T(x)=L(x)(x-8)-4</math>, then <math>L(4)=-8</math>. Therefore, it is possible to choose <math>T(x) = -8(x-8)-4 = -8x + 60</math>, so the goal is accomplished. As a reference, the polynomial we get is<br />
<br />
<cmath>P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315</cmath><br />
<cmath> = -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325</cmath><br />
<br />
== Critique of Critique of Critique ==<br />
<br />
The problem states the "least value" of <math>a</math>, so it is not needed to add the extra steps.<br />
<br />
== Critique of Critique of Critique of Critique==<br />
It is still necessary to show that the minimum is achievable.<br />
For example <math>x^2+1>0</math>, but <math>0</math> is not the least value of <math>x^2+1</math><br />
== Critique of all the previous Critiques ==<br />
Why are there so many?<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=B|num-b=24|after=Last question}}<br />
See also<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_25&diff=819612010 AMC 10B Problems/Problem 252016-12-21T23:35:18Z<p>Fuegocaliente: /* Critique of Critique of Critique of Critique */</p>
<hr />
<div>== Problem ==<br />
Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that<br />
<br />
<center><br />
<math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/><br />
<math>P(2) = P(4) = P(6) = P(8) = -a</math>.<br />
</center><br />
<br />
What is the smallest possible value of <math>a</math>?<br />
<br />
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math><br />
<br />
== Solution ==<br />
We observe that because <math>P(1) = P(3) = P(5) = P(7) = a</math>, if we define a new polynomial <math>R(x)</math> such that <math>R(x) = P(x) - a</math>, <math>R(x)</math> has roots when <math>P(x) = a</math>; namely, when <math>x=1,3,5,7</math>.<br />
<br />
Thus since <math>R(x)</math> has roots when <math>x=1,3,5,7</math>, we can factor the product <math>(x-1)(x-3)(x-5)(x-7)</math> out of <math>R(x)</math> to obtain a new polynomial <math>Q(x)</math> such that <math>(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a</math>.<br />
<br />
Then, plugging in values of <math>2,4,6,8,</math> we get <br />
<br />
<cmath>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</cmath><br />
<cmath>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a</cmath><br />
<cmath>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</cmath><br />
<cmath>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</cmath><br />
<br />
<math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</math><br />
Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>.<br />
Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.<br />
<br />
== Critique ==<br />
<br />
The above solution is incomplete. What is really proven is that 315 is a factor of <math>a</math>, if such an <math>a</math> exists. That only rules out answer A.<br />
<br />
To prove that the answer is correct, one could exhibit a polynomium that satisfies the requirements with <math>a=315</math>. Here's one: <math>P(x) = -8x^7 + 252 x^6 -3248x^5 + 22050x^4 -84392x^3 + 179928x^2-194592x+80325</math>.<br />
<br />
You get that from the <math>8\times8</math> matrix <math>M_{i,j} = i^{j-1}</math> and <math>y^T=(315,-315,\ldots,315,-315)</math> and computing <math>c=M^{-1}y</math> which comes out as the all-integer coefficients above.<br />
<br />
== Critique of Critique==<br />
<br />
First of all, the solution shows that <math>a</math> is a multiple of <math>315</math>, not a factor of <math>315</math>. Many people confuse the usage of the words 'factor' and 'multiple'. Secondly, even if <math>a</math> is a multiple of <math>315</math>, it does not mean that you can instantly get that the answer is <math>315</math> because we need to know that <math>a=315</math> is possible. After all, <math>a</math> is also a multiple of <math>1</math>, but <math>1</math> is definitely not the smallest possible number.<br />
<br />
To complete the solution, we can let <math>a = 315</math>, and then try to find <math>Q(x)</math>. We know from the above calculation that <math>Q(2)=42, Q(4)=-70, Q(6)=42</math>, and <math>Q(8)=-6</math>. Then we can let <math>Q(x) = T(x)(x-2)(x-6)+42</math>, getting <math>T(4)=28, T(8)=-4</math>. Let <math>T(x)=L(x)(x-8)-4</math>, then <math>L(4)=-8</math>. Therefore, it is possible to choose <math>T(x) = -8(x-8)-4 = -8x + 60</math>, so the goal is accomplished. As a reference, the polynomial we get is<br />
<br />
<cmath>P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315</cmath><br />
<cmath> = -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325</cmath><br />
<br />
== Critique of Critique of Critique ==<br />
<br />
The problem states the "least value" of <math>a</math>, so it is not needed to add the extra steps.<br />
<br />
== Critique of Critique of Critique of Critique==<br />
It is still necessary to show that the minimum is achievable.<br />
For example <math>x^2+1>0</math>, but <math>0</math> is not the least value of <math>x^2+1</math><br />
== Critique of all the previous Critiques ==<br />
Why are there so many?<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|ab=B|num-b=24|after=Last question}}<br />
Seealso<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Polyhedron&diff=81822Polyhedron2016-12-07T03:15:29Z<p>Fuegocaliente: /* Related figures */</p>
<hr />
<div>A polyhedron is a three-dimensional surface composed of at least four flat [[face]]s which encloses a region of [[space]]. These faces intersect in [[edge]]s and [[vertex|vertices]]. Polyhedra are 3-D analogues of [[polygon]]s. They can be thought of as sets of [[ordered]] triples.<br />
<br />
== Classification ==<br />
===Concavity===<br />
Polyhedra can be [[convex polyhedron | convex]] or [[concave polyhedron | concave]].<br />
<br />
===Number of sides===<br />
===Regular polyhedra===<br />
They have congruent faces, angles, and edges. Only regular tetrahedra, hexahedra ([[cube]]s), octahedra, dodecahedra, and icosahedra exist. (In addition, a [[sphere]] could be thought of a polyhedron with an infinite number of faces.)<br />
== Common polyhedra ==<br />
The polyhedra most commonly encountered include:<br />
* [[tetrahedron]] - 4 faces<br />
* [[hexahedron]] - 6 faces<br />
etc.<br />
<br />
[[Prism]]s and [[pyramid]]s can be polyhedra.<br />
<br />
== Surface area ==<br />
The [[surface area]] of a polyhedron is the sum of the areas of its sides.<br />
<br />
== Volume ==<br />
The [[volume]] of a certain polyhedron is defined as <math>(B)h</math>, where B is the area of the base of the polyhedron and h is the height to this base.<br />
<br />
== Angles ==<br />
<br />
== Related figures ==<br />
* Polyhedral solids are the union of a polyhedron and the space that it encloses.<br />
* [[Polygon]]s<br />
* Polytopes<br />
{{stub}}<br />
[[Category:geometry]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Tetrahedron&diff=81821Tetrahedron2016-12-07T03:08:14Z<p>Fuegocaliente: </p>
<hr />
<div>The '''tetrahedron''' (plural ''tetrahedra'') or ''triangular pyramid'' is the simplest [[polyhedron]]. Tetrahedra have four [[vertex|vertices]], four [[triangle | triangular]] [[face]]s and six [[edge]]s. Three faces and three edges meet at each vertex.<br />
<br />
Any four points chosen in space will be the vertices of a tetrahedron as long as they do not all lie on a single [[plane]].<br />
<br />
Regular tetrahedra, in which all edges have equal [[length]] and all faces are [[congruent]] [[equilateral triangle]]s, are one of the five types of [[Platonic solid]]s.<br />
<br />
<asy><br />
import three; <br />
currentprojection = orthographic(-1.2,-0.2,0.4); <br />
triple[] P = {(0,0,(2/3)^.5),(3^(-0.5),0,0),(-1/2/3^.5,1/2,0),(-1/2/3^.5,-1/2,0)}; <br />
void drawFrontFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7));<br />
/* fill(P[x] -- P[y] -- P[z] -- cycle, rgb(0.7,0.7,0.7)); */ <br />
} <br />
void drawBackFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linetype("2 6")); <br />
} <br />
drawFrontFace(0,3,2);drawBackFace(0,1,3);drawBackFace(0,2,3);drawBackFace(1,2,3); <br />
</asy><br />
<br />
The [[polyhedral dual]] of a tetrahedron is another tetrahedron.<br />
<br />
{{stub}}<br />
[[Category:Geometry]]<br />
[[Category:Platonic solids]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=81807AoPS Wiki:FAQ2016-12-06T03:55:26Z<p>Fuegocaliente: /* I need more time for my homework, what should I do? */</p>
<hr />
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==== What is the olympiad package? ====<br />
<br />
:See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&t=165767 here]. The package contains a set of commands useful for drawing diagrams related to [[:Category:Olympiad Geometry Problems|olympiad geometry problems]].<br />
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== AoPSWiki ==<br />
==== Is there a guide for wiki syntax? ====<br />
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==== 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<br />
*'''AoPS'''- Art of Problem Solving, the website you're on right now!<br />
*'''AIME'''- American Invitational Mathematics Examination<br />
*'''AMC'''- American Math Competitions<br />
*'''ATM'''- At the Moment<br />
*'''brb'''- Be right Back<br />
*'''BTW'''- By the way<br />
*'''CEMC''' - Centre for Mathematics and Computing<br />
*'''C&P or C+P''' - Counting and Probability or Contests and Programs<br />
*'''EBWOP'''- Editing by way of post<br />
*'''FTW'''- For the Win, a game on AoPS<br />
*'''gg'''- Good Game<br />
*'''gj'''- Good Job<br />
*'''glhf'''-Good Luck Have Fun<br />
*'''gtgbye''' - Got to go<br />
*'''ID(R)K'''-I Don't (Really) Know<br />
*'''iff'''-If and only if<br />
*'''IIRC'''- If I recall correctly<br />
*'''IMO'''- In my opinion (or International Math Olympiad, depending on context)<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 />
*'''MOP'''- Mathematical Olympiad (Summer) Program<br />
*'''MSM'''- Middle School Math<br />
*'''NT'''- Number Theory<br />
*'''OBC'''- Online by computer<br />
*'''OMG'''- Oh My Gosh.<br />
*'''OP'''- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering<br />
*'''QED'''- Quod erat demonstrandum, Latin for Which was to be proven; some English mathematicians use it as an acronym for Quite Elegantly Done<br />
*'''QS&A'''- Questions, Suggestions, and Announcements Forum<br />
*'''ro(t)fl''' - Rolling on the floor laughing<br />
*'''sa''' - sa<br />
*'''smh''' - Shaking my head<br />
*'''USA(J)MO'''- USA (Junior) Mathematical Olympiad<br />
*'''V/LA'''- Vacation or Long Absence/Limited Access<br />
*'''WLOG'''- Without loss of generality<br />
*'''wrt'''- With respect to<br />
*'''wtg''' - Way to go<br />
*'''tytia'''- Thank you, that is all<br />
*'''xD'''- Bursting Laugh<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 suppose 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 "Request Extension" 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.<br />
:Otherwise, you can send an email to extensions@aops.com with your username, class name and ID, if known, and reason for extension. Someone should get back to you within a couple days.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_22&diff=817522016 AMC 8 Problems/Problem 222016-12-02T20:01:28Z<p>Fuegocaliente: /* Solution */</p>
<hr />
<div>Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. What is the area of the "bat wings" (shaded area)?<br />
<asy><br />
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br />
draw((3,0)--(1,4)--(0,0));<br />
fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br />
fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br />
label("$A$",(3.05,4.2));<br />
label("$B$",(2,4.2));<br />
label("$C$",(1,4.2));<br />
label("$D$",(0,4.2));<br />
label("$E$", (0,-0.2));<br />
label("$F$", (3,-0.2));<br />
label("$1$", (0.5, 4), N);<br />
label("$1$", (1.5, 4), N);<br />
label("$1$", (2.5, 4), N);<br />
label("$4$", (3.2, 2), E);<br />
</asy><br />
<br />
<math>\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5</math><br />
<br />
==Solution==<br />
The area of trapezoid <math>CBFE</math> is <math>\frac{1+3}2\cdot 4=8</math>. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a <math>3:1</math> ratio, so the height of the larger one is <math>3,</math> while the height of the smaller one is <math>1.</math> Thus, their areas are <math>\frac12</math> and <math>\frac92</math>. Subtracting these areas from the trapezoid, we get <math>8-\frac12-\frac92 =\boxed3</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }3}</math>.<br />
{{AMC8 box|year=2016|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Hexagon&diff=80663Hexagon2016-10-17T17:48:09Z<p>Fuegocaliente: /* Regular hexagons */</p>
<hr />
<div>{{stub}}<br />
<br />
A '''hexagon''' is a [[polygon]] with six [[edge]]s and six [[vertex | vertices]]. <br />
<br />
<br />
==Regular hexagons==<br />
Each internal [[angle]] of a [[Regular polygon | regular]] hexagon measures 120 [[degree (geometry) | degrees]], so the sum of the angles is <math>720^{\circ}</math>.<br />
<br />
[[Area]]: <math>\frac{3s^2\sqrt{3}}{2}</math> Where <math>s</math> is the side length of the hexagon.<br />
<br />
[[Apothem]], or [[inradius]]: <math>\dfrac{s\sqrt{3}}{2}</math><br />
<br />
[[Circumradius]]: <math>s</math><br />
<br />
==See Also==<br />
<br />
* [[Hexagonal number]]<br />
<br />
[[Category:Definition]]<br />
<br />
[[Category:Geometry]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Chinese_Remainder_Theorem&diff=78415Chinese Remainder Theorem2016-05-04T21:37:20Z<p>Fuegocaliente: /* Theorem */</p>
<hr />
<div>The '''Chinese Remainder Theorem''' is a [[number theory | number theoretic]] result. It is one of the only [[theorem]]s named for an oriental person or place, due to the closed development of mathematics in the western world.<br />
== Theorem ==<br />
<br />
Formally stated, the Chinese Remainder Theorem is as follows:<br />
<br />
Let <math>m</math> be [[relatively prime]] to <math>n</math>. Then each [[residue class]] mod <math>mn</math> is equal to the [[intersection]] of a unique residue class mod <math>m</math> and a unique residue class mod <math>n</math>, and the intersection of each residue class mod <math>m</math> with a residue class mod <math>n</math> is a residue class mod <math>mn</math>.<br />
<br />
<br />
Simply stated:<br />
<br />
Suppose you wish to find the least number <math>x</math> which leaves a remainder of:<br />
<br />
<center><br />
<math> \begin{aligned} &y_{1} \text{ when divided by } &d_{1}\\<br />
&y_{2} \text{ when divided by } &d_{2}\\<br />
&\vdots &\vdots\\<br />
&y_{n} \text{ when divided by } & d_{n}\\ \end{aligned} </math><br />
</center><br />
<br />
such that <math>d_{1}</math> , <math>d_{2}</math> , ... <math>d_{n}</math> are all relatively prime. <br />
Let <math>M = d_{1}d_{2} \cdots d_{n}</math>, and <math>b_{i} = \frac{M}{d_{i}}</math>. <br />
Now if the numbers <math>a_{i}</math> satisfy:<br />
<center><br />
<math>a_{i}b_{i} - 1 \equiv 0 \pmod {d_{i}} </math><br />
</center><br />
for every <math>1 \leq i \leq n</math>, then a solution for <math>x</math> is:<br />
<center><br />
<math>x = \sum_{i=1}^n a_{i}b_{i}y_{i} \pmod M</math><br />
</center><br />
<br />
== Proof ==<br />
<br />
If <math>a \equiv b \pmod{mn}</math>, then <math>a</math> and <math>b</math> differ by a multiple of <math>mn</math>, so <math>a \equiv b \pmod{m}</math> and <math>a \equiv b \pmod{n}</math>. This is the first part of the theorem. The converse follows because <math>a</math> and <math>b</math> must differ by a multiple of <math>m</math> and <math>n</math>, and <math>\mbox{lcm}(m,n) = mn</math>. This is the second part of the theorem.<br />
<br />
== Applicability ==<br />
<br />
Much like the [[Fundamental Theorem of Arithmetic]], many people seem to take this theorem for granted before they consciously turn their attention to it. It ubiquity derives from the fact that many results can be easily proven mod (a power of a prime), and can then be generalized to mod <math>m</math> using the Chinese Remainder Theorem. For instance, [[Fermat's Little Theorem]] may be generalized to the [[Fermat-Euler Theorem]] in this manner.<br />
<br />
'''General Case''': the proof of the general case follows by induction to the above result (k-1) times.<br />
<br />
==Extended version of the theorem==<br />
Suppose one tried to divide a group of fish into <math>2</math>, <math>3</math> and <math>4</math> parts instead and found <math>1</math>, <math>1</math> and <math>2</math> fish left over, respectively. Any number with remainder <math>1</math> mod <math>2</math> must be [[odd integer | odd]] and any number with remainder <math>2</math> mod <math>4</math> must be [[even integer | even]]. Thus, the number of objects must be odd and even simultaneously, which is a contradiction. Thus, there must be restrictions on the numbers <math>a_1,\dots,a_n</math> to ensure that at least one solution exists. It follows that:<br />
<br />
''The solution exists if and only if <math>a_i\equiv a_j\mod \gcd(m_i,m_j)</math> for all <math>i,j</math> where <math>\gcd</math> stands for the [[greatest common divisor]]. Moreover, in the case when the problem is solvable, any two solutions differ by some [[common multiple]] of <math>m_1,\ldots,m_n</math>.'' (the extended version).<br />
<br />
==See Also==<br />
*[[Modular arithmetic/Introduction]]<br />
*[[Chicken McNugget Theorem]]<br />
<br />
==Discussion==<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80124&sid=df9439496046e0ff9f97cfc644408395 Here] is an AoPS thread in which the Chinese Remainder Theorem is discussed and implemented.<br />
<br />
[[Category:Number theory]]<br />
[[Category:Theorems]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Primorial&diff=77877Primorial2016-04-02T00:26:26Z<p>Fuegocaliente: </p>
<hr />
<div>Primorial is a the product of all the prime numbers less than or equal to itself. And n primorial is denoted as n#<br />
<br />
Examples:<br />
7# = 7*5*3*2 = 210<br />
<br />
8# is also <math>210</math> because 8 is not a prime.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Primorial&diff=77876Primorial2016-04-02T00:26:09Z<p>Fuegocaliente: </p>
<hr />
<div>Primorial is a the product of all the prime numbers less than or equal to itself. And n primorial is denoted as n#<br />
<br />
Examples:<br />
7# = 7*5*3*2 = 210<br />
8# is also <math>210</math> because 8 is not a prime.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Primorial&diff=77875Primorial2016-04-02T00:25:46Z<p>Fuegocaliente: </p>
<hr />
<div>Primorial is a the product of all the prime numbers less than or equal to itself. And n primorial is denoted as n#<br />
<br />
Examples:<br />
7# = 7\cdot5\cdot3\cdot2 = 210<br />
8# is also <math>210</math> because 8 is not a prime.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Primorial&diff=77874Primorial2016-04-02T00:25:08Z<p>Fuegocaliente: Created page with "Primorial is a the product of all the prime numbers less than or equal to itself. And n primorial is denoted as n# Examples: <cmath>7# = 7\cdot5\cdot3\cdot2 = 210</cmath> <ma..."</p>
<hr />
<div>Primorial is a the product of all the prime numbers less than or equal to itself. And n primorial is denoted as n#<br />
<br />
Examples:<br />
<cmath>7# = 7\cdot5\cdot3\cdot2 = 210</cmath><br />
<math>8#</math> is also <math>210</math> because 8 is not a prime.</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Combination&diff=72453Combination2015-10-12T22:42:42Z<p>Fuegocaliente: /* Formula */</p>
<hr />
<div>A '''combination''' is a way of choosing <math>r</math> objects from a set of <math>n</math> where the order in which the objects are chosen is irrelevant. We are generally concerned with finding the number of combinations of size <math>r</math> from an original set of size <math>n</math><br />
<br />
<br />
== Notation ==<br />
<br />
The common forms of denoting the number of combinations of <math>{r}</math> objects from a set of <math>{n}</math> objects is:<br />
<br />
* <math>\binom{n}{r}</math><br />
* <math>{C}(n,r)</math><br />
* <math>\,_{n} C_{r}</math><br />
* <math> C_n^{r} </math><br />
<br />
== Formula ==<br />
<br />
<math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math><br />
or<br />
<math>{{n}\choose{r}} = \frac{n*(n-1)}{r}</math><br />
<br />
== Derivation ==<br />
<br />
Consider the set of letters A, B, and C. There are <math>3!</math> different [[permutations]] of those letters. Since order doesn't matter with combinations, there is only one combination of those three.(<math>ABC,ACB,BAC,BCA,CAB,CBA</math> are all equivalent in combination but different in permutation.) <br />
<br />
In general, since for every permutation of <math>{r}</math> objects from <math>{n}</math> elements <math>P(n,r)</math> which is <math>\frac{n!}{n-r!}</math>.Now since in permutation the order of arrangement matters(<math>ABC</math> is not same as <math>ACB</math>) but in combinations the order of arrangement does not matter(<math>ABC</math> is equivalent to <math>ACB</math>).For its derivation see this [http://www.artofproblemsolving.com/Videos/external.php?video_id=181 video].<br />
<br />
Suppose <math>r</math> elements are selected out.For permutation <math>r</math> elements can be arranged in <math>r!</math> ways.We have overcounted the number of combinations by <math>r!</math> times.So We have <math>{r}!{C}({n},{r})=P(n,r)</math>, or <math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math>.<br />
<br />
==Formulas/Identities==<br />
* <math>\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{r}</math><br />
<br />
* <math>\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}</math><br />
<br />
* <math>\sum_{x=0}^{n} \binom{n}{x} = 2^n</math><br />
<br />
One of the many proofs is by first inserting <math>a = b = 1</math> into the [[binomial theorem]]. Because the combinations are the coefficients of <math>2^n</math>, and a and b disappear because they are 1, the sum is <math>2^n</math>.<br />
<br />
* <math>\sum_{i=0}^{k}\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}</math><br />
<br />
* <math>\binom{n}{r}=\binom{n}{n-r}</math><br />
<br />
We can prove this by putting the combinations in their algebraic form.<br />
<math>\binom{n}{n-r}=\frac{n!}{(n-r)!(n-(n-r))!}</math>. As we can see, <math>(n-(n-r))!=(n-n+r)=r!</math>. By the [[commutative property]], <math>\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}</math>. Because <math>\frac{n!}{r!(n-r)!}=\binom{n}{r}</math>, by the [[transitive property]], we can conclude that this is true for all non-negative integers n and r where n is greater than or equal to r. Another reason this is true is because that choosing what you don't want is the same as choosing what you do want, because by choosing what you don't want, you imply that you choose the rest. This identity is also the reason why [[Pascal's Triangle]] is symmetrical.<br />
<br />
== Examples ==<br />
<br />
* [[2005_AIME_II_Problems/Problem_1 | 2005 AIME II Problem 1]]<br />
<br />
== See also ==<br />
<br />
* [[Combinatorics]]<br />
* [[Combinatorial identity]]<br />
* [[Permutations]]<br />
* [[Pascal's Triangle]]<br />
* [[Generating function]]<br />
<br />
[[Category:Combinatorics]]<br />
[[Category:Definition]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Combination&diff=72346Combination2015-10-02T22:24:53Z<p>Fuegocaliente: /* Formula */</p>
<hr />
<div>A '''combination''' is a way of choosing <math>r</math> objects from a set of <math>n</math> where the order in which the objects are chosen is irrelevant. We are generally concerned with finding the number of combinations of size <math>r</math> from an original set of size <math>n</math><br />
<br />
<br />
== Notation ==<br />
<br />
The common forms of denoting the number of combinations of <math>{r}</math> objects from a set of <math>{n}</math> objects is:<br />
<br />
* <math>\binom{n}{r}</math><br />
* <math>{C}(n,r)</math><br />
* <math>\,_{n} C_{r}</math><br />
* <math> C_n^{r} </math><br />
<br />
== Formula ==<br />
<br />
<math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math><br />
or<br />
<math>\frac{n*(n-1)}{r}</math><br />
<br />
== Derivation ==<br />
<br />
Consider the set of letters A, B, and C. There are <math>3!</math> different [[permutations]] of those letters. Since order doesn't matter with combinations, there is only one combination of those three.(<math>ABC,ACB,BAC,BCA,CAB,CBA</math> are all equivalent in combination but different in permutation.) <br />
<br />
In general, since for every permutation of <math>{r}</math> objects from <math>{n}</math> elements <math>P(n,r)</math> which is <math>\frac{n!}{n-r!}</math>.Now since in permutation the order of arrangement matters(<math>ABC</math> is not same as <math>ACB</math>) but in combinations the order of arrangement does not matter(<math>ABC</math> is equivalent to <math>ACB</math>).For its derivation see this [http://www.artofproblemsolving.com/Videos/external.php?video_id=181 video].<br />
<br />
Suppose <math>r</math> elements are selected out.For permutation <math>r</math> elements can be arranged in <math>r!</math> ways.We have overcounted the number of combinations by <math>r!</math> times.So We have <math>{r}!{C}({n},{r})=P(n,r)</math>, or <math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math>.<br />
<br />
==Formulas/Identities==<br />
* <math>\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{r}</math><br />
<br />
* <math>\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}</math><br />
<br />
* <math>\sum_{x=0}^{n} \binom{n}{x} = 2^n</math><br />
<br />
One of the many proofs is by first inserting <math>a = b = 1</math> into the [[binomial theorem]]. Because the combinations are the coefficients of <math>2^n</math>, and a and b disappear because they are 1, the sum is <math>2^n</math>.<br />
<br />
* <math>\sum_{i=0}^{k}\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}</math><br />
<br />
* <math>\binom{n}{r}=\binom{n}{n-r}</math><br />
<br />
We can prove this by putting the combinations in their algebraic form.<br />
<math>\binom{n}{n-r}=\frac{n!}{(n-r)!(n-(n-r))!}</math>. As we can see, <math>(n-(n-r))!=(n-n+r)=r!</math>. By the [[commutative property]], <math>\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}</math>. Because <math>\frac{n!}{r!(n-r)!}=\binom{n}{r}</math>, by the [[transitive property]], we can conclude that this is true for all non-negative integers n and r where n is greater than or equal to r. Another reason this is true is because that choosing what you don't want is the same as choosing what you do want, because by choosing what you don't want, you imply that you choose the rest. This identity is also the reason why [[Pascal's Triangle]] is symmetrical.<br />
<br />
== Examples ==<br />
<br />
* [[2005_AIME_II_Problems/Problem_1 | 2005 AIME II Problem 1]]<br />
<br />
== See also ==<br />
<br />
* [[Combinatorics]]<br />
* [[Combinatorial identity]]<br />
* [[Permutations]]<br />
* [[Pascal's Triangle]]<br />
* [[Generating function]]<br />
<br />
[[Category:Combinatorics]]<br />
[[Category:Definition]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Combination&diff=72345Combination2015-10-02T22:24:17Z<p>Fuegocaliente: /* Derivation */</p>
<hr />
<div>A '''combination''' is a way of choosing <math>r</math> objects from a set of <math>n</math> where the order in which the objects are chosen is irrelevant. We are generally concerned with finding the number of combinations of size <math>r</math> from an original set of size <math>n</math><br />
<br />
<br />
== Notation ==<br />
<br />
The common forms of denoting the number of combinations of <math>{r}</math> objects from a set of <math>{n}</math> objects is:<br />
<br />
* <math>\binom{n}{r}</math><br />
* <math>{C}(n,r)</math><br />
* <math>\,_{n} C_{r}</math><br />
* <math> C_n^{r} </math><br />
<br />
== Formula ==<br />
<br />
<math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math><br />
<br />
== Derivation ==<br />
<br />
Consider the set of letters A, B, and C. There are <math>3!</math> different [[permutations]] of those letters. Since order doesn't matter with combinations, there is only one combination of those three.(<math>ABC,ACB,BAC,BCA,CAB,CBA</math> are all equivalent in combination but different in permutation.) <br />
<br />
In general, since for every permutation of <math>{r}</math> objects from <math>{n}</math> elements <math>P(n,r)</math> which is <math>\frac{n!}{n-r!}</math>.Now since in permutation the order of arrangement matters(<math>ABC</math> is not same as <math>ACB</math>) but in combinations the order of arrangement does not matter(<math>ABC</math> is equivalent to <math>ACB</math>).For its derivation see this [http://www.artofproblemsolving.com/Videos/external.php?video_id=181 video].<br />
<br />
Suppose <math>r</math> elements are selected out.For permutation <math>r</math> elements can be arranged in <math>r!</math> ways.We have overcounted the number of combinations by <math>r!</math> times.So We have <math>{r}!{C}({n},{r})=P(n,r)</math>, or <math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math>.<br />
<br />
==Formulas/Identities==<br />
* <math>\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{r}</math><br />
<br />
* <math>\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}</math><br />
<br />
* <math>\sum_{x=0}^{n} \binom{n}{x} = 2^n</math><br />
<br />
One of the many proofs is by first inserting <math>a = b = 1</math> into the [[binomial theorem]]. Because the combinations are the coefficients of <math>2^n</math>, and a and b disappear because they are 1, the sum is <math>2^n</math>.<br />
<br />
* <math>\sum_{i=0}^{k}\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}</math><br />
<br />
* <math>\binom{n}{r}=\binom{n}{n-r}</math><br />
<br />
We can prove this by putting the combinations in their algebraic form.<br />
<math>\binom{n}{n-r}=\frac{n!}{(n-r)!(n-(n-r))!}</math>. As we can see, <math>(n-(n-r))!=(n-n+r)=r!</math>. By the [[commutative property]], <math>\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}</math>. Because <math>\frac{n!}{r!(n-r)!}=\binom{n}{r}</math>, by the [[transitive property]], we can conclude that this is true for all non-negative integers n and r where n is greater than or equal to r. Another reason this is true is because that choosing what you don't want is the same as choosing what you do want, because by choosing what you don't want, you imply that you choose the rest. This identity is also the reason why [[Pascal's Triangle]] is symmetrical.<br />
<br />
== Examples ==<br />
<br />
* [[2005_AIME_II_Problems/Problem_1 | 2005 AIME II Problem 1]]<br />
<br />
== See also ==<br />
<br />
* [[Combinatorics]]<br />
* [[Combinatorial identity]]<br />
* [[Permutations]]<br />
* [[Pascal's Triangle]]<br />
* [[Generating function]]<br />
<br />
[[Category:Combinatorics]]<br />
[[Category:Definition]]</div>Fuegocalientehttps://artofproblemsolving.com/wiki/index.php?title=Combination&diff=72344Combination2015-10-02T22:22:59Z<p>Fuegocaliente: /* Derivation */</p>
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<div>A '''combination''' is a way of choosing <math>r</math> objects from a set of <math>n</math> where the order in which the objects are chosen is irrelevant. We are generally concerned with finding the number of combinations of size <math>r</math> from an original set of size <math>n</math><br />
<br />
<br />
== Notation ==<br />
<br />
The common forms of denoting the number of combinations of <math>{r}</math> objects from a set of <math>{n}</math> objects is:<br />
<br />
* <math>\binom{n}{r}</math><br />
* <math>{C}(n,r)</math><br />
* <math>\,_{n} C_{r}</math><br />
* <math> C_n^{r} </math><br />
<br />
== Formula ==<br />
<br />
<math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math><br />
<br />
== Derivation ==<br />
<br />
Consider the set of letters A, B, and C. There are <math>3!</math> different [[permutations]] of those letters. Since order doesn't matter with combinations, there is only one combination of those three.(<math>ABC,ACB,BAC,BCA,CAB,CBA</math> are all equivalent in combination but different in permutation.) <br />
<br />
In general, since for every permutation of <math>{r}</math> objects from <math>{n}</math> elements <math>P(n,r)</math> which is <math>\frac{n!}{n-r!}</math>.Now since in permutation the order of arrangement matters(<math>ABC</math> is not same as <math>ACB</math>) but in combinations the order of arrangement does not matter(<math>ABC</math> is equivalent to <math>ACB</math>).For its derivation see this [http://www.artofproblemsolving.com/Videos/external.php?video_id=181 video].<br />
<br />
Suppose <math>r</math> elements are selected out.For permutation <math>r</math> elements can be arranged in <math>r!</math> ways.We have overcounted the number of combinations by <math>r!</math> times.So We have <math>{r}!{C}({n},{r})=P(n,r)</math>, or <math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math>. Another way to compute the combination is <math>\frac{n*(n-1)}{r}</math><br />
<br />
==Formulas/Identities==<br />
* <math>\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{r}</math><br />
<br />
* <math>\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}</math><br />
<br />
* <math>\sum_{x=0}^{n} \binom{n}{x} = 2^n</math><br />
<br />
One of the many proofs is by first inserting <math>a = b = 1</math> into the [[binomial theorem]]. Because the combinations are the coefficients of <math>2^n</math>, and a and b disappear because they are 1, the sum is <math>2^n</math>.<br />
<br />
* <math>\sum_{i=0}^{k}\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}</math><br />
<br />
* <math>\binom{n}{r}=\binom{n}{n-r}</math><br />
<br />
We can prove this by putting the combinations in their algebraic form.<br />
<math>\binom{n}{n-r}=\frac{n!}{(n-r)!(n-(n-r))!}</math>. As we can see, <math>(n-(n-r))!=(n-n+r)=r!</math>. By the [[commutative property]], <math>\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}</math>. Because <math>\frac{n!}{r!(n-r)!}=\binom{n}{r}</math>, by the [[transitive property]], we can conclude that this is true for all non-negative integers n and r where n is greater than or equal to r. Another reason this is true is because that choosing what you don't want is the same as choosing what you do want, because by choosing what you don't want, you imply that you choose the rest. This identity is also the reason why [[Pascal's Triangle]] is symmetrical.<br />
<br />
== Examples ==<br />
<br />
* [[2005_AIME_II_Problems/Problem_1 | 2005 AIME II Problem 1]]<br />
<br />
== See also ==<br />
<br />
* [[Combinatorics]]<br />
* [[Combinatorial identity]]<br />
* [[Permutations]]<br />
* [[Pascal's Triangle]]<br />
* [[Generating function]]<br />
<br />
[[Category:Combinatorics]]<br />
[[Category:Definition]]</div>Fuegocaliente