https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ev3commander&feedformat=atomAoPS Wiki - User contributions [en]2022-01-27T07:37:52ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=89220Gmaas2017-12-28T01:00:28Z<p>Ev3commander: /* Known Facts About Gmaas */</p>
<hr />
<div>=== Known Facts About Gmaas ===<br />
-Gmass is Gmass.<br />
<br />
-Gmass is life.<br />
<br />
- sseraj once spelled gmaas as gmass by accident in Introduction to Geometry (1532).<br />
<br />
-He actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile<br />
<br />
-Gmass is a South Korean, North Korean, Palestini, Israeli, U.S., and Soviet citizen at the same time.<br />
<br />
-i am sand destroyed Gmass in FTW<br />
<br />
- sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br />
<br />
-Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten<br />
<br />
-Gmaas is roy moore's horse in the shape of a cat<br />
<br />
-Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over <math>289547987693</math> robux and <math>190348</math> in CPR.<br />
<br />
-This is all hypothetical EDIT: This is all factual <br />
<br />
- He is capable of salmon powers, according to PunSpark (ask him)<br />
<br />
The Gmaas told Richard Rusczyk to make AoPS<br />
<br />
-The Gmaas is everything. Yes you are part of the Gmaas-Dw789<br />
<br />
-The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999th dimension<br />
<br />
-Certain theories provide evidence that he IS darth plagueis the wise<br />
<br />
-Gmaas has met George Washington<br />
<br />
-Gmaas is "TIRED OF PEOPLE ADDING TO HIS PAGE!!" (Maas 45).<br />
<br />
-Gmaas has multiple accounts; some of them are pifinity, Lord_Baltimore, Spacehead1AU, and Electro3.0<br />
<br />
-Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br />
<br />
-Owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br />
<br />
- Made this page<br />
<br />
- King of the first men, the anduls<br />
<br />
-is a well known professor at MEOWston Academy<br />
<br />
-Gmass is a Tuna addict<br />
<br />
- won the reward of being cutest and (fattest) cat ever<br />
<br />
-Last sighting 1571-stretch-algebra-a 12/6/17<br />
<br />
- owner of sseraj not pet<br />
<br />
- embodiment of life and universe and beyond <br />
<br />
- Watches memes<br />
<br />
-After Death became the GOD OF HYPERDEATH and obtained over 9000 souls <br />
<br />
-Gmass's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br />
<br />
-Gmass is a certified Slytherin and probably the cutest cat ever.<br />
<br />
-Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br />
<br />
-Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br />
<br />
- Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br />
<br />
-Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br />
<br />
-Gmaas is a supreme overlord who must be given a miencraft DIAMOND<br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br />
<br />
- Gmaas is 5space's favorite animal. [https://www.youtube.com/watch?v=hSlb1ezRqfA&vl=en(Source)]<br />
<br />
- He lives with sseraj. <br />
<br />
-Gmaas is my favorite pokemon<br />
<br />
-Gmaas dislikes number theory but enjoys geometry.<br />
<br />
- Gmaas is cool<br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS</cmath> supercars, excluding the Purrari and the 138838383 Teslas. <br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
- He is distant relative of Mathcat1234<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks, because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
<br />
~Alligator Swamp A 1072 <br />
<br />
~Alligator Swamp B 1073<br />
<br />
~Introduction to Algebra A (1170)<br />
<br />
~Introduction to Algebra B (1529)<br />
<br />
~Welcome to Panda Town Gate 1076<br />
<br />
~Welcome to Gmaas Town Gate 1221<br />
<br />
~Welcome to Gmaas Town Gate 1125<br />
<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
<br />
~The other side of the ice in Antarctica<br />
<br />
~Feisty Alligator Swamp 1115<br />
<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
<br />
~Introduction to Counting and Probability 1142 <br />
<br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
<br />
~Intermediate Counting and Probability 1137<br />
<br />
~Intermediate Counting and Probability 1207<br />
<br />
~Posting student surveys<br />
<br />
~USF Castle Walls - Elven Tribe 1203<br />
<br />
~Dark Lord's Hut 1210<br />
<br />
~AMC 10 Problem Series 1200<br />
<br />
~Intermediate Number Theory 1138<br />
<br />
~Intermediate Number Theory 1476<br />
<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
<br />
~Algebra B 1112<br />
<br />
~Intermediate Algebra 1561 7:17 PM 12/11/16<br />
<br />
<br />
<br />
<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (Especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
EDIT: Nobody has yet seen him atop a tribal base yet.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: Gmaas rarely disguises himself as a penguin.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br />
<br />
-Aha!! An impostor!! <br />
http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br />
(look at the acronym).<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99999}{100000}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- EDIT: Gmaas is a he.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mew.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions, or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br />
https://artofproblemsolving.com/community/c287916h1291232<br />
<br />
- EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- EDIT: That has never happened and thus it does not contain the singularity of a black hole. <br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
-Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br />
<br />
- The original owner of Gmaas is gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyberhairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He is their symbol. Or he was, for one yoctosecond.<br />
EDIT: that is a lie. he IS THE Illuminati. april fools(it's actually july, i do april fools year round)<br />
<br />
- It has been wondered if gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmaal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks.<br />
<br />
- Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both gmaas and the DL (USF code name of the Dark Lord) love BBC. How gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br />
<br />
-The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br />
<br />
- EDIT: The above fact is somewhat irrelevant.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has <math>57843504</math> regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
-Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
-Gmaas has a secret hidden garden full of turnips under sseraj's house.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance in the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything on sight? Nobody knows. (Except him.)<br />
<br />
-Gmaas lives on Gallifrey and in Gotham City (he has sleepovers with Batman).<br />
<br />
-Gmaas is an excellent driver.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
-Donald Trump once sang "All Hail the Chief" to Gmaas, 3 days after being sworn in as US President.<br />
<br />
- Gmaas likess to talk with rrusczyk from time to time.<br />
<br />
- gmaas can shoot fire from his paws.<br />
<br />
- Gmaas is the reason why the USF has the longest thread on AoPS.<br />
<br />
- He (or she) is an avid watcher of the popular T.V. show "Bernie Sanders and the Gauntlet of DOOM"<br />
<br />
- sseraj, in 1521 Introduction to Number Theory, posted an image of gmaas after saying "Who wants to see 5space?" at around 5:16 PM Mountain Time, noting gmaas was "also 5space"<br />
<br />
-EDIT: he also did it in Introduction to Algebra A once<br />
<br />
- Gmaas is now my HD background on my Mac.<br />
<br />
- In 1521 Into to Number Theory, sseraj posted an image of a 5space gmaas fusion. (First sighting) <br />
<br />
- Also confirmed that gmaas doesn't like ketchup because it was the only food left the photo.<br />
<br />
- In 1447 Intro to Geometry, sseraj posted a picture of gmaas with a rubik's cube suggesting that gmaas's has an average solve time of <math>-GMAAS</math> seconds.<br />
<br />
-Gmass beat Superman in a fight with ease<br />
<br />
-Gmass was an admin of Roblox<br />
<br />
-Gmass traveled around the world,paying so much <math>MONEY</math> just to eat :D<br />
<br />
-Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br />
Summary:<br />
<br />
-When Gmaas subtracts <math>0.\overline{99}</math> from <math>1</math>, the difference is greater than <math>0</math>.<br />
<br />
-Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br />
<br />
-Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br />
<br />
-The results of the revival is top secret, and nobody knows what happened.<br />
<br />
-sseraj, in 1496 Prealgebra 2, said that gmaas is Santacat.<br />
<br />
-sseraj likes to post a picture of gmaas in every class he passes by.<br />
<br />
-sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Ender is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmass is now wandering space in search for a home.<br />
EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br />
<br />
-Gmaas is the lord of the pokemans<br />
<br />
-Gmass can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will<br />
<br />
-Picture of Gmass http://i.imgur.com/PP9xi.png<br />
<br />
-Known by Mike Miller<br />
<br />
-Gmaas got mad at sseraj once, so he locked him in his own freezer<br />
<br />
-Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br />
<br />
-Gmaas is an obviously omnipotent cat.<br />
<br />
-ehawk11 met him<br />
<br />
-sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br />
<br />
-sseraj has posted pictures of gmass in intro to algebra, before class started, with the title, "caption contest" anyone who posted a caption mysteriously vanished in the middle of the night. <br />
EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmass is typing this through his/her account...)<br />
<br />
- gmass has once slept on your bed and made it wet<br />
<br />
-It is rumored that rrusczyk is actually gmaas in disguise<br />
<br />
-Gmaas is suspected to be a Mewtwo in disguise<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- [s]Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".[/s]Sorry, this was a rick roll troll.<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath and themoocow are also writing a book about Gmaas<br />
<br />
-Ornx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br />
<br />
- gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br />
<br />
-oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br />
<br />
-Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br />
<br />
-No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br />
<br />
-In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br />
https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br />
https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br />
https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br />
https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br />
<br />
<br />
- Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br />
<br />
-<math>Another</math> sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br />
<br />
-Gmaas has been sighted several times on the Global Announcements forum<br />
<br />
-Gmaas uses the following transportation: [img]http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg[/img]</div>Ev3commanderhttps://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems/Problem_5&diff=886512016 AIME II Problems/Problem 52017-11-26T22:28:46Z<p>Ev3commander: splel</p>
<hr />
<div>Triangle <math>ABC_0</math> has a right angle at <math>C_0</math>. Its side lengths are pairwise relatively prime positive integers, and its perimeter is <math>p</math>. Let <math>C_1</math> be the foot of the altitude to <math>\overline{AB}</math>, and for <math>n \geq 2</math>, let <math>C_n</math> be the foot of the altitude to <math>\overline{C_{n-2}B}</math> in <math>\triangle C_{n-2}C_{n-1}B</math>. The sum <math>\sum_{i=1}^\infty C_{n-2}C_{n-1} = 6p</math>. Find <math>p</math>.<br />
<br />
==Solution 1==<br />
Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ac}{b}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}</math>. Multiplying by the denominator and expanding, the equation becomes <math>\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a</math>. Cancelling <math>6a</math> and multiplying by <math>c</math> yields <math>ab=6bc+6c^2-6a^2-6ab</math>, so <math>7ab = 6bc+6b^2</math> and <math>7a=6b+6c</math>. Checking for Pythagorean triples gives <math>13,84,</math> and <math>85</math>, so <math>p=13+84+85=\boxed{182}</math><br />
<br />
Solution modified/fixed from Shaddoll's solution.<br />
<br />
==Solution 2==<br />
We start by splitting the sum of all <math>C_{n-2}C_{n-1}</math> into two parts: those where <math>n-2</math> is odd and those where <math>n-2</math> is even.<br />
<br />
<br />
First consider the sum of the lengths of the segments for which <math>n-2</math> is odd for each <math>n\geq2</math>. The perimeters of these triangles can be expressed using <math>p</math> and ratios that result because of similar triangles. Considering triangles where <math>n-2</math> is odd, we find that the perimeter for each such <math>n</math> is <math>p\left(\frac{C_{n-1}C_{n}}{C_{0}B}\right)</math>. Thus, <br />
<br />
<math>p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B</math>.<br />
<br />
<br />
Simplifying,<br />
<br />
<math>\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B\left(6+\frac{C_{0}B}{p}\right)</math>. (1)<br />
<br />
<br />
Continuing with a similar process for the sum of the lengths of the segments for which <math>n-2</math> is even,<br />
<br />
<math>p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B</math>.<br />
<br />
<br />
Simplifying,<br />
<br />
<math>\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B\left(7-\frac{C_{0}B}{p}\right)</math>. (2)<br />
<br />
<br />
Adding (1) and (2) together, we find that <br />
<br />
<math>6p=13C_{0}B \Rightarrow p=\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \Rightarrow \frac{7C_{0}B}{6}=C_{0}A+AB \Rightarrow 7C_{0}B=6C_{0}A + 6AB</math>.<br />
<br />
Setting <math>a=C_{0}B</math>, <math>b=C_{0}A</math>, and <math>c=AB</math>, we can now proceed as in Shaddoll's solution, and our answer is <math>p=13+84+85=\boxed{182}</math>.<br />
<br />
Solution by brightaz<br />
<br />
== Solution 3 ==<br />
<asy><br />
size(10cm);<br />
<br />
// Setup<br />
pair A, B;<br />
pair C0, C1, C2, C3, C4, C5, C6, C7, C8;<br />
A = (5, 0);<br />
B = (0, 3);<br />
C0 = (0, 0);<br />
C1 = foot(C0, A, B);<br />
C2 = foot(C1, C0, B);<br />
C3 = foot(C2, C1, B);<br />
C4 = foot(C3, C2, B);<br />
C5 = foot(C4, C3, B);<br />
C6 = foot(C5, C4, B);<br />
C7 = foot(C6, C5, B);<br />
C8 = foot(C7, C6, B);<br />
<br />
// Labels<br />
label("$A$", A, SE);<br />
label("$B$", B, NW);<br />
label("$C_0$", C0, SW);<br />
label("$C_1$", C1, NE);<br />
label("$C_2$", C2, W);<br />
label("$C_3$", C3, NE);<br />
label("$C_4$", C4, W);<br />
<br />
// Drawings<br />
draw(A--B--C0--cycle);<br />
draw(C0--C1--C2, red);<br />
draw(C2--C3--C4--C5--C6--C7--C8);<br />
draw(C0--C2, blue);<br />
</asy><br />
<br />
Let <math>a = BC_0</math>, <math>b = AC_0 </math>, and <math>c = AB</math>.<br />
Note that the total length of the red segments in the figure above is equal to the length of the blue segment times <math>\frac{a+c}{b}</math>.<br />
<br />
The desired sum is equal to the total length of the infinite path <math>C_0 C_1 C_2 C_3 \cdots</math>, shown in red in the figure below.<br />
Since each of the triangles <math>\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots</math> on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times <math>\frac{a+c}{b}</math>.<br />
In other words, we have that <math>a\left(\frac{a+c}{b}\right) = 6p</math>.<br />
<br />
Guessing and checking Pythagorean triples reveals that <math>a = 84</math>, <math>b=13</math>, <math>c = 85</math>, and <math>p = a + b + c = \boxed{182}</math> satisfies this equation.<br />
<br />
<asy><br />
size(10cm);<br />
<br />
// Setup<br />
pair A, B;<br />
pair C0, C1, C2, C3, C4, C5, C6, C7, C8;<br />
A = (5, 0);<br />
B = (0, 3);<br />
C0 = (0, 0);<br />
C1 = foot(C0, A, B);<br />
C2 = foot(C1, C0, B);<br />
C3 = foot(C2, C1, B);<br />
C4 = foot(C3, C2, B);<br />
C5 = foot(C4, C3, B);<br />
C6 = foot(C5, C4, B);<br />
C7 = foot(C6, C5, B);<br />
C8 = foot(C7, C6, B);<br />
<br />
// Labels<br />
label("$A$", A, SE);<br />
label("$B$", B, NW);<br />
label("$C_0$", C0, SW);<br />
label("$a$", (B+C0)/2, W);<br />
label("$b$", (A+C0)/2, S);<br />
label("$c$", (A+B)/2, NE);<br />
<br />
// Drawings<br />
draw(A--B--C0--cycle);<br />
draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red);<br />
draw(C0--B, blue);<br />
</asy><br />
<br />
== See also ==<br />
{{AIME box|year=2016|n=II|num-b=4|num-a=6}}</div>Ev3commanderhttps://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_2&diff=850062017 AIME II Problems/Problem 22017-03-24T22:57:56Z<p>Ev3commander: /* Solution */</p>
<hr />
<div>==Problem==<br />
The teams <math>T_1</math>, <math>T_2</math>, <math>T_3</math>, and <math>T_4</math> are in the playoffs. In the semifinal matches, <math>T_1</math> plays <math>T_4</math>, and <math>T_2</math> plays <math>T_3</math>. The winners of those two matches will play each other in the final match to determine the champion. When <math>T_i</math> plays <math>T_j</math>, the probability that <math>T_i</math> wins is <math>\frac{i}{i+j}</math>, and the outcomes of all the matches are independent. The probability that <math>T_4</math> will be the champion is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution==<br />
There are two scenarios in which <math>T_4</math> wins. The first scenario is where <math>T_4</math> beats <math>T_1</math>, <math>T_3</math> beats <math>T_2</math>, and <math>T_4</math> beats <math>T_3</math>, and the second scenario is where <math>T_4</math> beats <math>T_1</math>, <math>T_2</math> beats <math>T_3</math>, and <math>T_4</math> beats <math>T_2</math>. Consider the first scenario. The probability <math>T_4</math> beats <math>T_1</math> is <math>\frac{4}{4+1}</math>, the probability <math>T_3</math> beats <math>T_2</math> is <math>\frac{3}{3+2}</math>, and the probability <math>T_4</math> beats <math>T_3</math> is <math>\frac{4}{4+3}</math>. Therefore the first scenario happens with probability <math>\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}</math>. Consider the second scenario. The probability <math>T_4</math> beats <math>T_1</math> is <math>\frac{4}{1+4}</math>, the probability <math>T_2</math> beats <math>T_3</math> is <math>\frac{2}{2+3}</math>, and the probability <math>T_4</math> beats <math>T_2</math> is <math>\frac{4}{4+2}</math>. Therefore the second scenario happens with probability <math>\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}</math>. By summing these two probabilities, the probability that <math>T_4</math> wins is <math>\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}+\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}</math>. Because this expression is equal to <math>\frac{256}{525}</math>, the answer is <math>256+525=\boxed{781}</math>.<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Ev3commanderhttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_18&diff=820682009 AMC 12A Problems/Problem 182016-12-30T01:32:57Z<p>Ev3commander: /* Alternate Solution */</p>
<hr />
<div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}}<br />
<br />
== Problem ==<br />
For <math>k > 0</math>, let <math>I_k = 10\ldots 064</math>, where there are <math>k</math> zeros between the <math>1</math> and the <math>6</math>. Let <math>N(k)</math> be the number of factors of <math>2</math> in the prime factorization of <math>I_k</math>. What is the maximum value of <math>N(k)</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10</math><br />
<br />
== Solution ==<br />
<br />
The number <math>I_k</math> can be written as <math>10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6</math>.<br />
<br />
For <math>k\in\{1,2,3\}</math> we have <math>I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)</math>. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have <math>N(k)=k+2\leq 5</math>.<br />
<br />
For <math>k>4</math> we have <math>I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)</math>. For <math>k>4</math> the value in the parentheses is odd, hence <math>N(k)=6</math>.<br />
<br />
This leaves the case <math>k=4</math>. We have <math>I_4 = 2^6 \left( 5^6 + 1 \right)</math>. The value <math>5^6 + 1</math> is obviously even. And as <math>5\equiv 1 \pmod 4</math>, we have <math>5^6 \equiv 1 \pmod 4</math>, and therefore <math>5^6 + 1 \equiv 2 \pmod 4</math>. Hence the largest power of <math>2</math> that divides <math>5^6+1</math> is <math>2^1</math>, and this gives us the desired maximum of the function <math>N</math>: <math>N(4) = \boxed{7}</math>.<br />
<br />
== Alternate Solution ==<br />
<br />
Notice that 2 is a prime factor of an integer <math>n</math> if and only if <math>n</math> is even. Therefore, given any sufficiently high positive integral value of <math>k</math>, dividing <math>I_k</math> by <math>2^6</math> yields a terminal digit of zero, and dividing by 2 again leaves us with <math>2^7 \cdot a = I_k</math> where <math>a</math> is an odd integer.<br />
Observe then that <math>\boxed{\textbf{(B)}7}</math> must be the maximum value for <math>N(k)</math> because whatever value we choose for <math>k</math>, <math>N(k)</math> must be less than or equal to <math>7</math>.<br />
<br />
EDIT: Isn't this solution incomplete because we need to show that <math>N(k) = 7</math> can be reached?<br />
<br />
An example of 7 being reached is 1000064. 1000064 divided by <math>2^7=128</math> is 7813.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2009|ab=A|num-b=17|num-a=19}}<br />
{{AMC10 box|year=2009|ab=A|num-b=24|after=Last Question}}<br />
{{MAA Notice}}</div>Ev3commander