https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ilikepi12&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T22:22:16ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=User:MRENTHUSIASM&diff=179916User:MRENTHUSIASM2022-11-01T13:06:00Z<p>Ilikepi12: /* User Count */</p>
<hr />
<div>==User Count==<br />
If this is your first time visiting this page, then edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">170</font></center><br />
<br />
==Introduction==<br />
Hi Everyone,<br />
<br />
I am a hedgehog who likes Message Board halping, funny pictures, emoji wars, and AoPS Wiki contributions. Keep these fun things coming!<br />
<br />
This hedgehog represents my attitude towards mathematics--victorious and enthusiastic!<br />
[[File:Commander Hedgehog.png|center]]<br />
And here is my personified self-portrait:<br />
[[File:Enthusiastic.gif|center]]<br />
~MRENTHUSIASM<br />
<br />
==AoPS Bio==<br />
Jerry graduated from Northeastern University in 2018 as a mathematics and computer science combined major. He has eleven years of experience working with students. He has served as an AoPS instructor and a message board moderator since 2019. During his last two years in high school, he served as the math club president and prepared lively lectures on contest math. As a result, he inspired many kids to get interested in math and participate in math competitions (such as AMC and ARML). For the last eight years, he has worked with Math League to write math contests for students from grades 4 through 12 in the USA. In his mind, nothing is more rewarding than educating tomorrow’s bright minds. In 2020 he published the book English-Chinese Mathematics Encyclopedia: Algebra, Geometry, and Pre-Calculus in Tsinghua University Press in China. In his spare time, he loves contributing to the AoPS Wiki and playing chess and all sorts of board games, especially Scotland Yard, and Samurai.<br />
<br />
==AoPS Wiki Contributions==<br />
Below are my contributions to the AMC/AIME Problems' Solutions in the AoPS Wiki. I understand that in this communal Wiki, we all have to collaborate and make compromises. <i><b>In any of my solutions, if you find flaws and/or want major revisions, then please contact me via my [[User_talk:MRENTHUSIASM|user talk page]] or the private messaging system before you take action. I am sure that we can work things out.</b></i><br />
<br />
Thank you for your cooperation, and hope you enjoy reading my solutions.<br />
<br />
Finally, [https://www.youtube.com/channel/UCR0u7fEppRjD-OFOEbtzhKw here] is my YouTube channel (MRENTHUSIASM) for the video solutions. Go ahead and subscribe--you know you want to. <math>\smiley{}</math> <br />
<br />
~MRENTHUSIASM<br />
<br />
===AJHSME / AMC 8===<br />
* [[2007_AMC_8_Problems/Problem_8|2007 AMC 8 Problem 8]] (Solutions 1, 2)<br />
* [[2007_AMC_8_Problems/Problem_10|2007 AMC 8 Problem 10]] (Solution)<br />
* [[2009_AMC_8_Problems/Problem_4|2009 AMC 8 Problem 4]] (Solution)<br />
* [[2017_AMC_8_Problems/Problem_24|2017 AMC 8 Problem 24]] (Solution 2)<br />
* [[2020_AMC_8_Problems/Problem_1|2020 AMC 8 Problem 1]] (Solutions 2, 3)<br />
* [[2020_AMC_8_Problems/Problem_3|2020 AMC 8 Problem 3]] (Solution 1)<br />
* [[2020_AMC_8_Problems/Problem_4|2020 AMC 8 Problem 4]] (Solutions 2, 4)<br />
* [[2022_AMC_8_Problems/Problem_3|2022 AMC 8 Problem 3]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_4|2022 AMC 8 Problem 4]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_5|2022 AMC 8 Problem 5]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_6|2022 AMC 8 Problem 6]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_8|2022 AMC 8 Problem 8]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_9|2022 AMC 8 Problem 9]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_10|2022 AMC 8 Problem 10]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_13|2022 AMC 8 Problem 13]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_14|2022 AMC 8 Problem 14]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_16|2022 AMC 8 Problem 16]] (Solutions 1, 2)<br />
* [[2022_AMC_8_Problems/Problem_18|2022 AMC 8 Problem 18]] (Solution 1)<br />
* [[2022_AMC_8_Problems/Problem_20|2022 AMC 8 Problem 20]] (Solution 2)<br />
* [[2022_AMC_8_Problems/Problem_22|2022 AMC 8 Problem 22]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_24|2022 AMC 8 Problem 24]] (Remark)<br />
* [[2022_AMC_8_Problems/Problem_25|2022 AMC 8 Problem 25]] (Solution 1)<br />
<br />
===AMC 10===<br />
* [[2004_AMC_10A_Problems/Problem_24|2004 AMC 10A Problem 24]] (Solutions 1, 2)<br />
* [[2007_AMC_10A_Problems/Problem_20|2007 AMC 10A Problem 20]] (Solutions 1, 2, 3, 4, 5, 6, 7)<br />
* [[2009_AMC_10B_Problems/Problem_1|2009 AMC 10B Problem 1]] (Solutions 1, 2, 3)<br />
* [[2012_AMC_10B_Problems/Problem_10|2012 AMC 10B Problem 10]] (Solutions 1, 2)<br />
* [[2018_AMC_10A_Problems/Problem_1|2018 AMC 10A Problem 1]] (Solution)<br />
* [[2018_AMC_10A_Problems/Problem_5|2018 AMC 10A Problem 5]] (Solution 1)<br />
* [[2018_AMC_10A_Problems/Problem_7|2018 AMC 10A Problem 7]] (Solutions 1, 3)<br />
* [[2018_AMC_10A_Problems/Problem_8|2018 AMC 10A Problem 8]] (Solution 3)<br />
* [[2018_AMC_10A_Problems/Problem_17|2018 AMC 10A Problem 17]] (Solution 4)<br />
* [[2018_AMC_10A_Problems/Problem_21|2018 AMC 10A Problem 21]] (Remark)<br />
* [[2018_AMC_10A_Problems/Problem_23|2018 AMC 10A Problem 23]] (Solution 1)<br />
* [[2018_AMC_10A_Problems/Problem_25|2018 AMC 10A Problem 25]] (Solutions 1, 3)<br />
* [[2018_AMC_10B_Problems/Problem_2|2018 AMC 10B Problem 2]] (Solutions 1, 2)<br />
* [[2018_AMC_10B_Problems/Problem_15|2018 AMC 10B Problem 15]] (Solutions 1, 3)<br />
* [[2018_AMC_10B_Problems/Problem_19|2018 AMC 10B Problem 19]] (Solution 1)<br />
* [[2018_AMC_10B_Problems/Problem_20|2018 AMC 10B Problem 20]] (Solutions 1, 2)<br />
* [[2018_AMC_10B_Problems/Problem_21|2018 AMC 10B Problem 21]] (Solutions 1, 2)<br />
* [[2018_AMC_10B_Problems/Problem_24|2018 AMC 10B Problem 24]] (Diagram, Solution 5)<br />
* [[2020_AMC_10A_Problems/Problem_3|2020 AMC 10A Problem 3]] (Solutions 1, 2)<br />
* [[2020_AMC_10A_Problems/Problem_5|2020 AMC 10A Problem 5]] (Solution 3)<br />
* [[2020_AMC_10A_Problems/Problem_11|2020 AMC 10A Problem 11]] (Solution 5)<br />
* [[2020_AMC_10A_Problems/Problem_17|2020 AMC 10A Problem 17]] (Solution 1)<br />
* [[2020_AMC_10A_Problems/Problem_24|2020 AMC 10A Problem 24]] (Solution 8)<br />
* [[2020_AMC_10A_Problems/Problem_25|2020 AMC 10A Problem 25]] (Solution 2)<br />
* [[2020_AMC_10B_Problems/Problem_3|2020 AMC 10B Problem 3]] (Solution 1)<br />
* [[2020_AMC_10B_Problems/Problem_4|2020 AMC 10B Problem 4]] (Solution 3)<br />
* [[2020_AMC_10B_Problems/Problem_8|2020 AMC 10B Problem 8]] (Solutions 1, 2)<br />
* [[2020_AMC_10B_Problems/Problem_13|2020 AMC 10B Problem 13]] (Solution 1)<br />
* [[2020_AMC_10B_Problems/Problem_15|2020 AMC 10B Problem 15]] (Solutions 1, 3)<br />
* [[2020_AMC_10B_Problems/Problem_23|2020 AMC 10B Problem 23]] (Solution 1)<br />
* [[2021_AMC_10A_Problems/Problem_1|2021 AMC 10A Problem 1]] (Solution 2)<br />
* [[2021_AMC_10A_Problems/Problem_2|2021 AMC 10A Problem 2]] (Solutions 1, 2, 3, 4)<br />
* [[2021_AMC_10A_Problems/Problem_3|2021 AMC 10A Problem 3]] (Solution 3)<br />
* [[2021_AMC_10A_Problems/Problem_4|2021 AMC 10A Problem 4]] (Solutions 1, 3)<br />
* [[2021_AMC_10A_Problems/Problem_5|2021 AMC 10A Problem 5]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_10A_Problems/Problem_6|2021 AMC 10A Problem 6]] (Solutions 1, 2)<br />
* [[2021_AMC_10A_Problems/Problem_7|2021 AMC 10A Problem 7]] (Solutions 1, 3)<br />
* [[2021_AMC_10A_Problems/Problem_8|2021 AMC 10A Problem 8]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_10A_Problems/Problem_9|2021 AMC 10A Problem 9]] (Solution 2)<br />
* [[2021_AMC_10A_Problems/Problem_10|2021 AMC 10A Problem 10]] (Solution 7)<br />
* [[2021_AMC_10A_Problems/Problem_11|2021 AMC 10A Problem 11]] (Solutions 1, 3)<br />
* [[2021_AMC_10A_Problems/Problem_12|2021 AMC 10A Problem 12]] (Solution 1)<br />
* [[2021_AMC_10A_Problems/Problem_13|2021 AMC 10A Problem 13]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_10A_Problems/Problem_17|2021 AMC 10A Problem 17]] (Diagram, Solutions 1, 2)<br />
* [[2021_AMC_10A_Problems/Problem_18|2021 AMC 10A Problem 18]] (Solutions 1, 4)<br />
* [[2021_AMC_10A_Problems/Problem_20|2021 AMC 10A Problem 20]] (Solutions 2, 3)<br />
* [[2021_AMC_10A_Problems/Problem_21|2021 AMC 10A Problem 21]] (Diagram, Solution, Video Solution)<br />
* [[2021_AMC_10A_Problems/Problem_22|2021 AMC 10A Problem 22]] (Solution 2, Video Solution)<br />
* [[2021_AMC_10A_Problems/Problem_23|2021 AMC 10A Problem 23]] (Solution 3, Video Solution)<br />
* [[2021_AMC_10A_Problems/Problem_24|2021 AMC 10A Problem 24]] (Diagram, Solutions 1, 2, 4, Video Solution)<br />
* [[2021_AMC_10A_Problems/Problem_25|2021 AMC 10A Problem 25]] (Solutions 2, 3, Video Solution)<br />
* [[2021_AMC_10B_Problems/Problem_6|2021 AMC 10B Problem 6]] (Solution 2)<br />
* [[2021_AMC_10B_Problems/Problem_7|2021 AMC 10B Problem 7]] (Solution)<br />
* [[2021_AMC_10B_Problems/Problem_8|2021 AMC 10B Problem 8]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_10B_Problems/Problem_17|2021 AMC 10B Problem 17]] (Solution)<br />
* [[2021_AMC_10B_Problems/Problem_23|2021 AMC 10B Problem 23]] (Diagram)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_1|2021 Fall AMC 10A Problem 1]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_2|2021 Fall AMC 10A Problem 2]] (Solution)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_3|2021 Fall AMC 10A Problem 3]] (Solutions 1, 2, 3)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_4|2021 Fall AMC 10A Problem 4]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_6|2021 Fall AMC 10A Problem 6]] (Solution)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_7|2021 Fall AMC 10A Problem 7]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_10|2021 Fall AMC 10A Problem 10]] (Solution)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_11|2021 Fall AMC 10A Problem 11]] (Solution 3)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_12|2021 Fall AMC 10A Problem 12]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_15|2021 Fall AMC 10A Problem 15]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_16|2021 Fall AMC 10A Problem 16]] (Solutions 2, 3)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_17|2021 Fall AMC 10A Problem 17]] (Diagrams, Solution 5)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_18|2021 Fall AMC 10A Problem 18]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_19|2021 Fall AMC 10A Problem 19]] (Diagram)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_20|2021 Fall AMC 10A Problem 20]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_21|2021 Fall AMC 10A Problem 21]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_22|2021 Fall AMC 10A Problem 22]] (Diagram)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_24|2021 Fall AMC 10A Problem 24]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_25|2021 Fall AMC 10A Problem 25]] (Solution 2)<br />
* [[2021_Fall_AMC_10B_Problems/Problem_2|2021 Fall AMC 10B Problem 2]] (Solutions 1, 2)<br />
* [[2021_Fall_AMC_10B_Problems/Problem_4|2021 Fall AMC 10B Problem 4]] (Solutions 1, 2)<br />
* [[2021_Fall_AMC_10B_Problems/Problem_7|2021 Fall AMC 10B Problem 7]] (Solution 1)<br />
<br />
===AHSME / AMC 12===<br />
* [[1976_AHSME_Problems/Problem_24|1976 AHSME Problem 24]] (Solution)<br />
* [[1976_AHSME_Problems/Problem_25|1976 AHSME Problem 25]] (Solution)<br />
* [[1976_AHSME_Problems/Problem_27|1976 AHSME Problem 27]] (Solutions 1, 2)<br />
* [[1976_AHSME_Problems/Problem_28|1976 AHSME Problem 28]] (Solution)<br />
* [[1976_AHSME_Problems/Problem_30|1976 AHSME Problem 30]] (Solution)<br />
* [[1978_AHSME_Problems/Problem_20|1978 AHSME Problem 20]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_19|1982 AHSME Problem 19]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_21|1982 AHSME Problem 21]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_23|1982 AHSME Problem 23]] (Solutions 1, 2)<br />
* [[1982_AHSME_Problems/Problem_27|1982 AHSME Problem 27]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_28|1982 AHSME Problem 28]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_29|1982 AHSME Problem 29]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_30|1982 AHSME Problem 30]] (Solution)<br />
* [[1990_AHSME_Problems/Problem_26|1990 AHSME Problem 26]] (Solutions 1, 2)<br />
* [[1991_AHSME_Problems/Problem_6|1991 AHSME Problem 6]] (Solution)<br />
* [[1991_AHSME_Problems/Problem_20|1991 AHSME Problem 20]] (Solution)<br />
* [[1991_AHSME_Problems/Problem_27|1991 AHSME Problem 27]] (Solution)<br />
* [[2004_AMC_12A_Problems/Problem_16|2004 AMC 12A Problem 16]] (Solution)<br />
* [[2004_AMC_12A_Problems/Problem_17|2004 AMC 12A Problem 17]]: same as [[2004_AMC_10A_Problems/Problem_24|2004 AMC 10A Problem 24]] (Solutions 1, 2)<br />
* [[2009_AMC_12B_Problems/Problem_1|2009 AMC 12B Problem 1]]: same as [[2009_AMC_10B_Problems/Problem_1|2009 AMC 10B Problem 1]] (Solutions 1, 2, 3)<br />
* [[2014_AMC_12A_Problems/Problem_18|2014 AMC 12A Problem 18]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_2|2018 AMC 12A Problem 2]] (Solutions 1, 2)<br />
* [[2018_AMC_12A_Problems/Problem 4|2018 AMC 12A Problem 4]]: same as [[2018_AMC_10A_Problems/Problem_5|2018 AMC 10A Problem 5]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_7|2018 AMC 12A Problem 7]]: same as [[2018_AMC_10A_Problems/Problem_7|2018 AMC 10A Problem 7]] (Solutions 1, 3)<br />
* [[2018_AMC_12A_Problems/Problem_12|2018 AMC 12A Problem 12]]: same as [[2018_AMC_10A_Problems/Problem_17|2018 AMC 10A Problem 17]] (Solution 4)<br />
* [[2018_AMC_12A_Problems/Problem_14|2018 AMC 12A Problem 14]] (Solutions 1, 2, 3, 4, 5)<br />
* [[2018_AMC_12A_Problems/Problem_16|2018 AMC 12A Problem 16]]: same as [[2018_AMC_10A_Problems/Problem_21|2018 AMC 10A Problem 21]] (Remark)<br />
* [[2018_AMC_12A_Problems/Problem_17|2018 AMC 12A Problem 17]]: same as [[2018_AMC_10A_Problems/Problem_23|2018 AMC 10A Problem 23]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_19|2018 AMC 12A Problem 19]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_21|2018 AMC 12A Problem 21]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_22|2018 AMC 12A Problem 22]] (Solutions 1, 2, 3)<br />
* [[2018_AMC_12A_Problems/Problem_23|2018 AMC 12A Problem 23]] (Diagram, Solution 2)<br />
* [[2018_AMC_12A_Problems/Problem_24|2018 AMC 12A Problem 24]] (Solutions 1, 2, 3)<br />
* [[2018_AMC_12A_Problems/Problem_25|2018 AMC 12A Problem 25]]: same as [[2018_AMC_10A_Problems/Problem_25|2018 AMC 10A Problem 25]] (Solutions 1, 3)<br />
* [[2018_AMC_12B_Problems/Problem_2|2018 AMC 12B Problem 2]]: same as [[2018_AMC_10B_Problems/Problem_2|2018 AMC 10B Problem 2]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_4|2018 AMC 12B Problem 4]] (Solution)<br />
* [[2018_AMC_12B_Problems/Problem_6|2018 AMC 12B Problem 6]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_8|2018 AMC 12B Problem 8]] (Solution 1)<br />
* [[2018_AMC_12B_Problems/Problem_9|2018 AMC 12B Problem 9]] (Solutions 1, 2, 3, 4, 5, 6)<br />
* [[2018_AMC_12B_Problems/Problem_11|2018 AMC 12B Problem 11]]: same as [[2018_AMC_10B_Problems/Problem_15|2018 AMC 10B Problem 15]] (Solutions 1, 3)<br />
* [[2018_AMC_12B_Problems/Problem_12|2018 AMC 12B Problem 12]] (Solution)<br />
* [[2018_AMC_12B_Problems/Problem_13|2018 AMC 12B Problem 13]] (Solutions 1, 2, 3)<br />
* [[2018_AMC_12B_Problems/Problem_14|2018 AMC 12B Problem 14]]: same as [[2018_AMC_10B_Problems/Problem_19|2018 AMC 10B Problem 19]] (Solution 1)<br />
* [[2018_AMC_12B_Problems/Problem_15|2018 AMC 12B Problem 15]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_16|2018 AMC 12B Problem 16]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_17|2018 AMC 12B Problem 17]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_18|2018 AMC 12B Problem 18]]: same as [[2018_AMC_10B_Problems/Problem_20|2018 AMC 10B Problem 20]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_19|2018 AMC 12B Problem 19]]: same as [[2018_AMC_10B_Problems/Problem_21|2018 AMC 10B Problem 21]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_20|2018 AMC 12B Problem 20]]: same as [[2018_AMC_10B_Problems/Problem_24|2018 AMC 10B Problem 24]] (Diagram, Solution 5)<br />
* [[2018_AMC_12B_Problems/Problem_21|2018 AMC 12B Problem 21]] (Diagram, Solution)<br />
* [[2018_AMC_12B_Problems/Problem_22|2018 AMC 12B Problem 22]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_23|2018 AMC 12B Problem 23]] (Diagram, Solutions 1, 2, 3)<br />
* [[2020_AMC_12A_Problems/Problem_1|2020 AMC 12A Problem 1]] (Solution 3)<br />
* [[2020_AMC_12A_Problems/Problem_6|2020 AMC 12A Problem 6]] (Solution 2)<br />
* [[2020_AMC_12A_Problems/Problem_8|2020 AMC 12A Problem 8]]: same as [[2020_AMC_10A_Problems/Problem_11|2020 AMC 10A Problem 11]] (Solution 5)<br />
* [[2020_AMC_12A_Problems/Problem_9|2020 AMC 12A Problem 9]] (Solution)<br />
* [[2020_AMC_12A_Problems/Problem_10|2020 AMC 12A Problem 10]] (Solutions 1, 2, 5)<br />
* [[2020_AMC_12A_Problems/Problem_15|2020 AMC 12A Problem 15]] (Solutions 1, 2)<br />
* [[2020_AMC_12A_Problems/Problem_23|2020 AMC 12A Problem 23]]: same as [[2020_AMC_10A_Problems/Problem_25|2020 AMC 10A Problem 25]] (Solution 2)<br />
* [[2020_AMC_12A_Problems/Problem_25|2020 AMC 12A Problem 25]] (Solution 1, Remark)<br />
* [[2020_AMC_12B_Problems/Problem_3|2020 AMC 12B Problem 3]]: same as [[2020_AMC_10B_Problems/Problem_3|2020 AMC 10B Problem 3]] (Solution 1)<br />
* [[2020_AMC_12B_Problems/Problem_4|2020 AMC 12B Problem 4]]: same as [[2020_AMC_10B_Problems/Problem_4|2020 AMC 10B Problem 4]] (Solution 3)<br />
* [[2020_AMC_12B_Problems/Problem_5|2020 AMC 12B Problem 5]] (Solution 1)<br />
* [[2020_AMC_12B_Problems/Problem_6|2020 AMC 12B Problem 6]] (Solution 2)<br />
* [[2020_AMC_12B_Problems/Problem_10|2020 AMC 12B Problem 10]] (Diagram, Solution 2)<br />
* [[2020_AMC_12B_Problems/Problem_12|2020 AMC 12B Problem 12]] (Diagram)<br />
* [[2020_AMC_12B_Problems/Problem_13|2020 AMC 12B Problem 13]] (Solutions 1, 4)<br />
* [[2020_AMC_12B_Problems/Problem_19|2020 AMC 12B Problem 19]]: same as [[2020_AMC_10B_Problems/Problem_23|2020 AMC 10B Problem 23]] (Solution 1)<br />
* [[2021_AMC_12A_Problems/Problem_2|2021 AMC 12A Problem 2]] (Solutions 3, 4)<br />
* [[2021_AMC_12A_Problems/Problem_3|2021 AMC 12A Problem 3]]: same as [[2021_AMC_10A_Problems/Problem_3|2021 AMC 10A Problem 3]] (Solution 3)<br />
* [[2021_AMC_12A_Problems/Problem_4|2021 AMC 12A Problem 4]]: same as [[2021_AMC_10A_Problems/Problem_7|2021 AMC 10A Problem 7]] (Solutions 1, 3)<br />
* [[2021_AMC_12A_Problems/Problem_5|2021 AMC 12A Problem 5]]: same as [[2021_AMC_10A_Problems/Problem_8|2021 AMC 10A Problem 8]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_6|2021 AMC 12A Problem 6]] (Solutions 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_7|2021 AMC 12A Problem 7]]: same as [[2021_AMC_10A_Problems/Problem_9|2021 AMC 10A Problem 9]] (Solution 2)<br />
* [[2021_AMC_12A_Problems/Problem_8|2021 AMC 12A Problem 8]] (Solution)<br />
* [[2021_AMC_12A_Problems/Problem_9|2021 AMC 12A Problem 9]]: same as [[2021_AMC_10A_Problems/Problem_10|2021 AMC 10A Problem 10]] (Solution 7)<br />
* [[2021_AMC_12A_Problems/Problem_10|2021 AMC 12A Problem 10]]: same as [[2021_AMC_10A_Problems/Problem_12|2021 AMC 10A Problem 12]] (Solution 1)<br />
* [[2021_AMC_12A_Problems/Problem_11|2021 AMC 12A Problem 11]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_13|2021 AMC 12A Problem 13]] (Solutions 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_14|2021 AMC 12A Problem 14]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_15|2021 AMC 12A Problem 15]] (Solutions 1, 2, 4)<br />
* [[2021_AMC_12A_Problems/Problem_17|2021 AMC 12A Problem 17]]: same as [[2021_AMC_10A_Problems/Problem_17|2021 AMC 10A Problem 17]] (Diagram, Solutions 1, 2)<br />
* [[2021_AMC_12A_Problems/Problem_18|2021 AMC 12A Problem 18]]: same as [[2021_AMC_10A_Problems/Problem_18|2021 AMC 10A Problem 18]] (Solutions 1, 4)<br />
* [[2021_AMC_12A_Problems/Problem_19|2021 AMC 12A Problem 19]] (Solutions 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_21|2021 AMC 12A Problem 21]] (Solution 2, Video Solution)<br />
* [[2021_AMC_12A_Problems/Problem_22|2021 AMC 12A Problem 22]] (Solutions 1, 2)<br />
* [[2021_AMC_12A_Problems/Problem_23|2021 AMC 12A Problem 23]]: same as [[2021_AMC_10A_Problems/Problem_23|2021 AMC 10A Problem 23]] (Solution 3, Video Solution)<br />
* [[2021_AMC_12A_Problems/Problem_24|2021 AMC 12A Problem 24]] (Solution 1)<br />
* [[2021_AMC_12A_Problems/Problem_25|2021 AMC 12A Problem 25]] (Solution 1)<br />
* [[2021_AMC_12B_Problems/Problem_4|2021 AMC 12B Problem 4]]: same as [[2021_AMC_10B_Problems/Problem_6|2021 AMC 10B Problem 6]] (Solution 2)<br />
* [[2021_AMC_12B_Problems/Problem_11|2021 AMC 12B Problem 11]] (Solution 4)<br />
* [[2021_AMC_12B_Problems/Problem_14|2021 AMC 12B Problem 14]] (Solution 1)<br />
* [[2021_AMC_12B_Problems/Problem_20|2021 AMC 12B Problem 20]] (Solution 1)<br />
* [[2021_AMC_12B_Problems/Problem_21|2021 AMC 12B Problem 21]] (Solution 2)<br />
* [[2021_AMC_12B_Problems/Problem_23|2021 AMC 12B Problem 23]] (Solution 4)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_1|2021 Fall AMC 12A Problem 1]]: same as [[2021_Fall_AMC_10A_Problems/Problem_1|2021 Fall AMC 10A Problem 1]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_2|2021 Fall AMC 12A Problem 2]]: same as [[2021_Fall_AMC_10A_Problems/Problem_2|2021 Fall AMC 10A Problem 2]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_3|2021 Fall AMC 12A Problem 3]]: same as [[2021_Fall_AMC_10A_Problems/Problem_4|2021 Fall AMC 10A Problem 4]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_5|2021 Fall AMC 12A Problem 5]]: same as [[2021_Fall_AMC_10A_Problems/Problem_6|2021 Fall AMC 10A Problem 6]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_6|2021 Fall AMC 12A Problem 6]]: same as [[2021_Fall_AMC_10A_Problems/Problem_7|2021 Fall AMC 10A Problem 7]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_7|2021 Fall AMC 12A Problem 7]]: same as [[2021_Fall_AMC_10A_Problems/Problem_10|2021 Fall AMC 10A Problem 10]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_9|2021 Fall AMC 12A Problem 9]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_10|2021 Fall AMC 12A Problem 10]]: same as [[2021_Fall_AMC_10A_Problems/Problem_12|2021 Fall AMC 10A Problem 12]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_12|2021 Fall AMC 12A Problem 12]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_13|2021 Fall AMC 12A Problem 13]] (Diagram, Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_16|2021 Fall AMC 12A Problem 16]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_17|2021 Fall AMC 12A Problem 17]]: same as [[2021_Fall_AMC_10A_Problems/Problem_20|2021 Fall AMC 10A Problem 20]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_18|2021 Fall AMC 12A Problem 18]]: same as [[2021_Fall_AMC_10A_Problems/Problem_21|2021 Fall AMC 10A Problem 21]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_23|2021 Fall AMC 12A Problem 23]]: same as [[2021_Fall_AMC_10A_Problems/Problem_25|2021 Fall AMC 10A Problem 25]] (Solution 2)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_24|2021 Fall AMC 12A Problem 24]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_25|2021 Fall AMC 12A Problem 25]] (Solution 1)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_2|2021 Fall AMC 12B Problem 2]]: same as [[2021_Fall_AMC_10B_Problems/Problem_2|2021 Fall AMC 10B Problem 2]] (Solutions 1, 2)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_3|2021 Fall AMC 12B Problem 3]]: same as [[2021_Fall_AMC_10B_Problems/Problem_4|2021 Fall AMC 10B Problem 4]] (Solutions 1, 2)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_5|2021 Fall AMC 12B Problem 5]]: same as [[2021_Fall_AMC_10B_Problems/Problem_7|2021 Fall AMC 10B Problem 7]] (Solution 1)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_10|2021 Fall AMC 12B Problem 10]] (Solution)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_12|2021 Fall AMC 12B Problem 12]] (Solutions 1, 3)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_17|2021 Fall AMC 12B Problem 17]] (Solution 4)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_18|2021 Fall AMC 12B Problem 18]] (Solution 1)<br />
<br />
===AIME===<br />
* [[1984_AIME_Problems/Problem_4|1984 AIME Problem 4]] (Solutions 1, 2)<br />
* [[1984_AIME_Problems/Problem_10|1984 AIME Problem 10]] (Solution 3)<br />
* [[1985_AIME_Problems/Problem_12|1985 AIME Problem 12]] (Solutions 1, 2, 3)<br />
* [[1986_AIME_Problems/Problem_2|1986 AIME Problem 2]] (Solutions 1, 2)<br />
* [[1987_AIME_Problems/Problem_14|1987 AIME Problem 14]] (Solutions 1, 2, 3)<br />
* [[1988_AIME_Problems/Problem_8|1988 AIME Problem 8]] (Solution 1)<br />
* [[1988_AIME_Problems/Problem_13|1988 AIME Problem 13]] (Solution 6)<br />
* [[1989_AIME_Problems/Problem_1|1989 AIME Problem 1]] (Solutions 3, 5)<br />
* [[1989_AIME_Problems/Problem_8|1989 AIME Problem 8]] (Solutions 1, 2, 3)<br />
* [[1989_AIME_Problems/Problem_9|1989 AIME Problem 9]] (Solutions 1, 2, 4)<br />
* [[1990_AIME_Problems/Problem_8|1990 AIME Problem 8]] (Solution, Remark)<br />
* [[1992_AIME_Problems/Problem_6|1992 AIME Problem 6]] (Solution 1)<br />
* [[1993_AIME_Problems/Problem_7|1993 AIME Problem 7]] (Solution 1)<br />
* [[1993_AIME_Problems/Problem_8|1993 AIME Problem 8]] (Solutions 3, 5)<br />
* [[2019_AIME_II_Problems/Problem_7|2019 AIME II Problem 7]] (Diagram)<br />
* [[2020_AIME_I_Problems/Problem_5|2020 AIME I Problem 5]] (Solution 9)<br />
* [[2021_AIME_I_Problems/Problem_1|2021 AIME I Problem 1]] (Solution 1)<br />
* [[2021_AIME_I_Problems/Problem_2|2021 AIME I Problem 2]] (Solution 5)<br />
* [[2021_AIME_I_Problems/Problem_3|2021 AIME I Problem 3]] (Solution 3)<br />
* [[2021_AIME_I_Problems/Problem_7|2021 AIME I Problem 7]] (Remark)<br />
* [[2021_AIME_I_Problems/Problem_8|2021 AIME I Problem 8]] (Solution 1, Remark)<br />
* [[2021_AIME_I_Problems/Problem_9|2021 AIME I Problem 9]] (Diagram, Solution 1)<br />
* [[2021_AIME_I_Problems/Problem_10|2021 AIME I Problem 10]] (Solution 2)<br />
* [[2021_AIME_I_Problems/Problem_11|2021 AIME I Problem 11]] (Diagram, Solution 1)<br />
* [[2021_AIME_I_Problems/Problem_12|2021 AIME I Problem 12]] (Solution)<br />
* [[2021_AIME_I_Problems/Problem_14|2021 AIME I Problem 14]] (Solutions 2, 3)<br />
* [[2021_AIME_I_Problems/Problem_15|2021 AIME I Problem 15]] (Diagram, Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_1|2021 AIME II Problem 1]] (Solution 3, Remark)<br />
* [[2021_AIME_II_Problems/Problem_2|2021 AIME II Problem 2]] (Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_3|2021 AIME II Problem 3]] (Solution 2)<br />
* [[2021_AIME_II_Problems/Problem_4|2021 AIME II Problem 4]] (Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_5|2021 AIME II Problem 5]] (Solutions 2, 5)<br />
* [[2021_AIME_II_Problems/Problem_6|2021 AIME II Problem 6]] (Solution 3)<br />
* [[2021_AIME_II_Problems/Problem_7|2021 AIME II Problem 7]] (Solution 3)<br />
* [[2021_AIME_II_Problems/Problem_8|2021 AIME II Problem 8]] (Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_9|2021 AIME II Problem 9]] (Solution, Remarks)<br />
* [[2021_AIME_II_Problems/Problem_10|2021 AIME II Problem 10]] (Diagram, Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_11|2021 AIME II Problem 11]] (Solution 2)<br />
* [[2021_AIME_II_Problems/Problem_12|2021 AIME II Problem 12]] (Solution 2)<br />
* [[2021_AIME_II_Problems/Problem_13|2021 AIME II Problem 13]] (Solutions 1, 3)<br />
* [[2021_AIME_II_Problems/Problem_14|2021 AIME II Problem 14]] (Diagram, Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_15|2021 AIME II Problem 15]] (Solution 2)<br />
* [[2022_AIME_I_Problems/Problem_1|2022 AIME I Problem 1]] (Solutions 1, 2, Video Solution)<br />
* [[2022_AIME_I_Problems/Problem_2|2022 AIME I Problem 2]] (Solution 1, Video Solution)<br />
* [[2022_AIME_I_Problems/Problem_3|2022 AIME I Problem 3]] (Diagram, Video Solution)<br />
* [[2022_AIME_I_Problems/Problem_4|2022 AIME I Problem 4]] (Solution 1, Video Solution)<br />
* [[2022_AIME_I_Problems/Problem_5|2022 AIME I Problem 5]] (Solution 2)<br />
* [[2022_AIME_I_Problems/Problem_7|2022 AIME I Problem 7]] (Solution)<br />
* [[2022_AIME_I_Problems/Problem_8|2022 AIME I Problem 8]] (Diagram)<br />
<br />
==AoPS Forums==<br />
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Let's have a good time in this forum. Feel free to discuss math classes, math competitions, AoPS Wiki contributions, funny pictures, emojis, and ... HEDGEHOGS! <math>\smiley{}</math><br />
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[[File:Hedgehog LALALALALA.gif|center]]<br />
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==Publications==<br />
<i><b>English-Chinese Mathematics Encyclopedia: Algebra, Geometry, and Pre-Calculus</b></i> (Tsinghua University Press in Beijing, China)<br />
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[[File:English-Chinese Mathematics Encyclopedia.jpeg|600px|center]]<br />
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==Artworks==<br />
In this section, I will show you my artworks of certificates and posters. Hope you enjoy them! <math>\smiley{}</math><br />
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I especially thank [[User:Flamekhoemberish|Flame Kho]] for teaching me how to use the Chuangkit design platform.<br />
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===Hall of Fame in MATHCOUNTS/AMC 8 Basics (2557)===<br />
[[File:2021 AMC 8 Perfect Scorers.png|center|750px]] <p><br />
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[[File:2557 Gold.png|center|750px]] <p><br />
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[[File:2557 Bronze.png|center|750px]]<br />
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===Hall of Fame in MATHCOUNTS/AMC 8 Basics (2786)===<br />
[[File:2846 2022 AMC 8 Perfect Scorers.png|center|500px]] <p><br />
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[[File:2846 Gold.png|center|750px]] <p><br />
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[[File:2846 Silver.png|center|750px]] <p><br />
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[[File:Funny Guy.png|center|750px]]<br />
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==Cats==<br />
As you might know, I am particularly fond of cats. My family has raised more than ten cats. These little angels are so adorable! MEOW MEOW MEOW!!! 🐱🐱🐱<br />
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===Luna===<br />
[[File:Luna.jpg|center|300px]]<br />
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===Kidney Bean===<br />
<center>[[File:Kidney Bean 1.jpg|300px]]&nbsp;&nbsp;&nbsp;[[File:Kidney Bean 2.jpg|300px]]</center><br />
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[[File:Jin Bao.jpg|center|300px]]<br />
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[[File:Little Gray Gray 1.jpg|center|300px]] <p><br />
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==External Links==<br />
* [[User_talk:MRENTHUSIASM|User Talk Page]] <br> Welcome to my user talk page here. Feel free to leave any message you like.<br />
* [[User:Flamekhoemberish|Flame Kho's User Page]] <br> Recently I am helping my friend Flame Kho to create his user page. Enjoy his AMC Solutions and artworks! <math>\smiley{}</math></div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_5&diff=1777162022 AIME II Problems/Problem 52022-08-30T17:49:11Z<p>Ilikepi12: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Twenty distinct points are marked on a circle and labeled <math>1</math> through <math>20</math> in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original <math>20</math> points.<br />
<br />
==Solution==<br />
<br />
Let <math>a</math>, <math>b</math>, and <math>c</math> be the vertex of a triangle that satisfies this problem, where <math>a > b > c</math>.<br />
<cmath>a - b = p_1</cmath><br />
<cmath>b - c = p_2</cmath><br />
<cmath>a - c = p_3</cmath><br />
<br />
<math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.<br />
<br />
Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{072}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_5&diff=1777152022 AIME II Problems/Problem 52022-08-30T17:47:54Z<p>Ilikepi12: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Twenty distinct points are marked on a circle and labeled <math>1</math> through <math>20</math> in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original <math>20</math> points.<br />
<br />
==Solution==<br />
<br />
Let <math>a</math>, <math>b</math>, and <math>c</math> be the vertex of a triangle that satisfies this problem, where <math>a > b > c</math>.<br />
<cmath>a - b = p_1</cmath><br />
<cmath>b - c = p_2</cmath><br />
<cmath>a - c = p_3</cmath><br />
<br />
<math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.<br />
<br />
Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{072}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
Note: I believe that you also need <math>20 \geq a</math>, so you cannot just simply do <math>18 \cdot 4</math>. In addition, it is possible for <cmath>b - c = 2</cmath> <cmath>a - b = p_1</cmath> to occur. You would need to do casework to make sure <math>a</math> does not go above 20. <br />
<br />
If you did the casework, it should come out to be <math>(15+13+7+1) \cdot 2 = \boxed{{072}}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_5&diff=1777142022 AIME II Problems/Problem 52022-08-30T17:47:06Z<p>Ilikepi12: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Twenty distinct points are marked on a circle and labeled <math>1</math> through <math>20</math> in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original <math>20</math> points.<br />
<br />
==Solution==<br />
<br />
Let <math>a</math>, <math>b</math>, and <math>c</math> be the vertex of a triangle that satisfies this problem, where <math>a > b > c</math>.<br />
<cmath>a - b = p_1</cmath><br />
<cmath>b - c = p_2</cmath><br />
<cmath>a - c = p_3</cmath><br />
<br />
<math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.<br />
<br />
Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{072}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
Note: I believe that you also need <math>20 \geq a</math>, so you cannot just simply do <math>18 \cdot 4</math>. In addition, it is possible for <cmath>b - c = 2</cmath> <cmath>a - b = p_1</cmath> to occur. You would need to do casework to make sure <math>a</math> does not go above 20. <br />
<br />
If you did the casework, it should come out to be <math>(15+13+7+1) \cdot 2 = \boxed{{072}}</math>. It just so happens that <math>72 = 18 \cdot 4</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_5&diff=1777132022 AIME II Problems/Problem 52022-08-30T17:46:50Z<p>Ilikepi12: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Twenty distinct points are marked on a circle and labeled <math>1</math> through <math>20</math> in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original <math>20</math> points.<br />
<br />
==Solution==<br />
<br />
Let <math>a</math>, <math>b</math>, and <math>c</math> be the vertex of a triangle that satisfies this problem, where <math>a > b > c</math>.<br />
<cmath>a - b = p_1</cmath><br />
<cmath>b - c = p_2</cmath><br />
<cmath>a - c = p_3</cmath><br />
<br />
<math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.<br />
<br />
Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{072}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
Note: I believe that you also need <math>20 \geq a</math>, so you cannot just simply do <math>18 \cdot 4</math>. In addition, it is possible for <cmath>b - c = 2</cmath> <cmath>a - b = p_1</cmath> to occur. You would need to do casework to make sure <math>a</math> does not go above 20. <br />
<br />
If you did the casework, it should come out to be <math>(15+13+7+1) \cdot 2 = \boxed{\textbf{072}}</math>. It just so happens that <math>72 = 18 \cdot 4</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_5&diff=1777122022 AIME II Problems/Problem 52022-08-30T17:45:08Z<p>Ilikepi12: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Twenty distinct points are marked on a circle and labeled <math>1</math> through <math>20</math> in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original <math>20</math> points.<br />
<br />
==Solution==<br />
<br />
Let <math>a</math>, <math>b</math>, and <math>c</math> be the vertex of a triangle that satisfies this problem, where <math>a > b > c</math>.<br />
<cmath>a - b = p_1</cmath><br />
<cmath>b - c = p_2</cmath><br />
<cmath>a - c = p_3</cmath><br />
<br />
<math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.<br />
<br />
Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{072}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
Note: I believe that you also need <math>20 \geq a</math>, so you cannot just simply do <math>18 \cdot 4</math>. In addition, it is possible for <cmath>b - c = 2</cmath> <cmath>a - b = p_1</cmath> to occur. You would need to do casework to make sure <math>a</math> does not go above 20.<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_5&diff=1777112022 AIME II Problems/Problem 52022-08-30T17:44:32Z<p>Ilikepi12: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Twenty distinct points are marked on a circle and labeled <math>1</math> through <math>20</math> in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original <math>20</math> points.<br />
<br />
==Solution==<br />
<br />
Let <math>a</math>, <math>b</math>, and <math>c</math> be the vertex of a triangle that satisfies this problem, where <math>a > b > c</math>.<br />
<cmath>a - b = p_1</cmath><br />
<cmath>b - c = p_2</cmath><br />
<cmath>a - c = p_3</cmath><br />
<br />
<math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.<br />
<br />
Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{072}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
Note: I believe that you also need <math>20 >= a</math>, so you cannot just simply do 18*4. In addition, it is possible for <cmath>b - c = 2</cmath> <cmath>a - b = p_1</cmath> to occur. You would need to do casework to make sure <math>a</math> does not go above 20.<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_5&diff=1777102022 AIME II Problems/Problem 52022-08-30T17:44:09Z<p>Ilikepi12: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Twenty distinct points are marked on a circle and labeled <math>1</math> through <math>20</math> in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original <math>20</math> points.<br />
<br />
==Solution==<br />
<br />
Let <math>a</math>, <math>b</math>, and <math>c</math> be the vertex of a triangle that satisfies this problem, where <math>a > b > c</math>.<br />
<cmath>a - b = p_1</cmath><br />
<cmath>b - c = p_2</cmath><br />
<cmath>a - c = p_3</cmath><br />
<br />
<math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.<br />
<br />
Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{072}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
Note: I believe that you also need <math>20 >= a</math>, so you cannot just simply do 18*4. In addition, it is possible for <cmath>b - c = 2</cmath>, and <cmath>a - b = p_1</cmath>. You would need to do casework to make sure <math>a</math> does not go above 20.<br />
<br />
==See Also==<br />
{{AIME box|year=2022|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=176626User:Piphi2022-08-03T18:55:47Z<p>Ilikepi12: /* User Count */</p>
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<div>{{User:Piphi/Template:Header}}<br />
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<br />
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</div></div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_21&diff=1426912009 AMC 10B Problems/Problem 212021-01-18T23:54:51Z<p>Ilikepi12: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
What is the remainder when <math>3^0 + 3^1 + 3^2 + \cdots + 3^{2009}</math> is divided by 8?<br />
<br />
<math>\mathrm{(A)}\ 0\qquad<br />
\mathrm{(B)}\ 1\qquad<br />
\mathrm{(C)}\ 2\qquad<br />
\mathrm{(D)}\ 4\qquad<br />
\mathrm{(E)}\ 6</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
The sum of any four consecutive powers of 3 is divisible by <math>3^0 + 3^1 + 3^2 +3^3 = 40</math> and hence is divisible by 8. Therefore<br />
: <math>(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{2009})</math><br />
is divisible by 8. So the required remainder is <math>3^0 + 3^1 = \boxed {4}</math>. The answer is <math>\mathrm{(D)}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
We have <math>3^2 = 9 \equiv 1 \pmod 8</math>. Hence for any <math>k</math> we have <math>3^{2k}\equiv 1^k = 1 \pmod 8</math>, and then <math>3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3 \pmod 8</math>.<br />
<br />
Therefore our sum gives the same remainder modulo <math>8</math> as <math>1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3</math>. There are <math>2010</math> terms in the sum, hence there are <math>2010/2 = 1005</math> pairs <math>1+3</math>, and thus the sum is <math>1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
We have the formula <math>\frac{a(r^n-1)}{r-1}</math> for the sum of a finite geometric sequence which we want to find the residue modulo 8.<br />
<cmath>\frac{1 \cdot (3^{2010}-1)}{2}</cmath><br />
<cmath>\frac{3^{2010}-1}{2} = \frac{9^{1005}-1}{2}</cmath><br />
<cmath>\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8</cmath><br />
Therefore, the numerator of the fraction is divisible by <math>8</math>. However, when we divide the numerator by <math>2</math>, we get a remainder of <math>4</math> modulo <math>8</math>, giving us <math>\mathrm{(D)}</math>.<br />
<br />
Note: you need to prove that <cmath>\frac{9^{1005}-1}{2}</cmath> is not congruent to 0 mod 16 because if so, then the whole thing would be congruent to 0 mod 8, even after dividing by 2 ~ ilikepi12<br />
<br />
== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Ilikepi12https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_22&diff=1411822013 AMC 10A Problems/Problem 222020-12-31T22:54:03Z<p>Ilikepi12: /* Video Solution by Richard Rusczyk */</p>
<hr />
<div>==Problem==<br />
<br />
Six spheres of radius <math>1</math> are positioned so that their centers are at the vertices of a regular hexagon of side length <math>2</math>. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br />
<br />
<br />
<math> \textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2 </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution 1==<br />
<br />
Set up an isosceles triangle between the center of the <math>8</math>th sphere and two opposite ends of the hexagon. Then set up another triangle between the point of tangency of the <math>7</math>th and <math>8</math>th spheres, and the points of tangency between the <math>7</math>th sphere and <math>2</math> of the original spheres on opposite sides of the hexagon. Express each side length of the triangles in terms of <math>r</math> (the radius of sphere <math>8</math>) and <math>h</math> (the height of the first triangle). You can then use Pythagorean Theorem to set up two equations for the two triangles, and find the values of <math>h</math> and <math>r</math>.<br />
<br />
<math>(1+r)^2=2^2+h^2</math><br />
<br />
<math>(3\sqrt{2})^2=3^2+(h+r)^2</math><br />
<br />
<math>r = \boxed{\textbf{(B) }\frac{3}{2}}</math><br />
<br />
==Solution 2==<br />
<br />
We have a regular hexagon with side length <math>2</math> and six spheres on each vertex with radius <math>1</math> that are internally tangent, therefore, drawing radii to the tangent points would create this regular hexagon.<br />
<br />
Imagine a 2D overhead view. There is a larger sphere which the <math>6</math> spheres are internally tangent to, with the center in the center of the hexagon. To find the radius of the larger sphere we must first, either by prior knowledge or by deducing from the angle sum that the hexagon can be split into <math>6</math> equilateral triangles from its vertices, that the radius is two more than the small radius, or <math>3</math>. <br />
<br />
<br />
<asy><br />
draw(circle((0.5,0.866025404),0.5));<br />
draw(circle((-0.5,0.866025404),0.5));<br />
draw(circle((1,0),0.5));<br />
draw(circle((-1,0),0.5));<br />
draw(circle((0.5,-0.866025404),0.5));<br />
draw(circle((-0.5,-0.866025404),0.5));<br />
draw(circle((0,0),1.5));<br />
<br />
draw((-0.5,0.866025404)--(0.5,0.866025404));<br />
draw((-1,0)--(1,0));<br />
draw((-0.5,-0.866025404)--(0.5,-0.866025404));<br />
<br />
draw((-1,0)--(-0.5,0.866025404));<br />
draw((-0.5,-0.866025404)--(0.5,0.866025404));<br />
draw((0.5,-0.866025404)--(1,0));<br />
<br />
draw((0.5,0.866025404)--(1,0));<br />
draw((-0.5,0.866025404)--(0.5,-0.866025404));<br />
draw((-1,0)--(-0.5,-0.866025404));<br />
</asy><br />
<br />
diagram made by erics118<br />
<br />
<br />
Now imagine the figure in <math>3</math> dimensions. The 8th sphere must be sitting atop of the <math>6</math> spheres, which is the only possibility for it to be tangent to all the <math>6</math> small spheres externally and the larger sphere internally. The ring of <math>6</math> small spheres is symmetrical and the 8th sphere will be resting atop it with its center aligned with the diameter of the large sphere.<br />
<br />
We can now create a right triangle with one leg being the line from a vertex of the hexagon to the center of the hexagon and the hypotenuse being the line from the center of the 7th sphere to the center of the 8th sphere. Let the radius of our 8th sphere be <math>r</math>. As previously mentioned, the distance from the center of the hexagon to one of its vertices is <math>2</math>. The distance between the centers will be <math>3-r</math>. The hypotenuse will be <math>r+1</math>. <br />
<br />
We now have a right triangle. Applying the Pythagorean Theorem, <math>2^2+(3-r)^2=(1+r)^2</math>. Expanding and solving for <math>r</math>, we find <math>r=\frac{12}{8}=\boxed{\textbf{(B)}\frac{3}{2}}</math>.<br />
<br />
==Video Solution by Richard Rusczyk==<br />
https://www.youtube.com/watch?v=83JgSTi_0VE <br />
<br />
hint: turn a 3D geometry problem into a 2D problem by taking cross-sections! Much easier to visualize.<br />
<br />
What type of cross-section...? Well, that's in the video :D<br />
<br />
<br />
~BakedPotato66 and dolphin7<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2013|ab=A|num-b=21|num-a=23}}<br />
{{AMC12 box|year=2013|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Ilikepi12