https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Rbhale12&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T21:58:00ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/August_16,_2011&diff=41538AoPS Wiki talk:Problem of the Day/August 16, 20112011-08-16T23:12:05Z<p>Rbhale12: Created page with "We know <math>{2011\choose0}+{2011\choose1}+...+{2011\choose2011}=2^{2011}</math>. The quantity we need to find is simply: <math>{2011\choose0}+{2011\choose1}+...+{2011\choose..."</p>
<hr />
<div>We know <math>{2011\choose0}+{2011\choose1}+...+{2011\choose2011}=2^{2011}</math>.<br />
<br />
The quantity we need to find is simply: <br />
<br />
<math>{2011\choose0}+{2011\choose1}+...+{2011\choose2011}-{2011\choose0}-{2011\choose2011}</math><br />
<br />
<math>2^{2011}-{2011\choose0}-{2011\choose2011} \Rightarrow 2^{2011}-1-1=\boxed{2^{2011}-2}</math>.</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_3&diff=414831991 AIME Problems/Problem 32011-08-15T13:33:56Z<p>Rbhale12: </p>
<hr />
<div>== Problem ==<br />
Expanding <math>(1+0.2)^{1000}_{}</math> by the binomial theorem and doing no further manipulation gives<br />
<br />
<math>{1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}</math><br />
<math>= A_0 + A_1 + A_2 + \cdots + A_{1000},</math><br />
where <math>A_k = {1000 \choose k}(0.2)^k</math> for <math>k = 0,1,2,\ldots,1000</math>. For which <math>k_{}^{}</math> is <math>A_k^{}</math> the largest?<br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Let <math>0<x_{}^{}<1</math>. Then we may write <math>A_{k}^{}={N\choose k}x^{k}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>. Taking logarithms in both sides of this last equation and using the well-known fact <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^{},b_{}^{}>0</math>), we have<br />
<br />
<math><br />
\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\log\left[\prod_{j=1}^{k}\frac{(N-j+1)x}{j}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .<br />
</math><br />
<br />
Now, <math>\log(A_{k}^{})</math> keeps increasing with <math>k_{}^{}</math> as long as the arguments <math>\frac{(N-j+1)x}{j}>1</math> in each of the <math>\log\big[\;\big]</math> terms (recall that <math>\log y_{}^{} <0</math> if <math>0<y_{}^{}<1</math>). Therefore, the integer <math>k_{}^{}</math> that we are looking for must satisfy <math>k=\Big\lfloor\frac{(N+1)x}{1+x}\Big\rfloor</math>, where <math>\lfloor z_{}^{}\rfloor</math> denotes the greatest integer less than or equal to <math>z_{}^{}</math>. <br />
<br />
In summary, substituting <math>N_{}^{}=1000</math> and <math>x_{}^{}=0.2</math> we finally find that <math>k_{}^{}=166</math>.<br />
<br />
===Solution 2===<br />
<br />
We know that once we have found the largest value of <math>k</math>, all values after <math>A_k</math> are less than <math>A_k</math>. Therefore, we are looking for the largest possible value such that:<br />
<br />
<math>{\frac{1}{5}}^k\cdot {{1000} \choose {k}}>{\frac{1}{5}}^{k+1}\cdot {{1000} \choose {k+1}}</math><br />
<br />
Dividing by <math>{\frac{1}{5}}^k</math> gives:<br />
<br />
<math>{1000\choose k}>{\frac{1}{5}}\cdot{1000\choose {k+1}}</math>.<br />
<br />
We can express these binomial coefficients as factorials.<br />
<br />
<math>\frac{1000!}{(1000-k)!\cdot(k)!}>{\frac{1}{5}}\cdot\frac{1000!}{(1000-k-1)!\cdot{(k+1)!}}</math><br />
<br />
We note that the <math>1000!</math> can cancel. Also, <math>(1000-k)!=(1000-k)(1000-k-1)!</math>. Similarly, <math>(k+1)!=(k+1)k!</math>.<br />
<br />
Canceling these terms yields,<br />
<br />
<math>\frac{1}{(1000-k)}>{\frac{1}{5}}\cdot\frac{1}{(k+1)}</math><br />
<br />
Cross multiplying gives:<br />
<br />
<math>5k+1>1000-k \Rightarrow k>166.5</math><br />
<br />
Therefore, since this identity holds for all values of <math>k>166.5</math>, the largest possible value of <math>k</math> is <math>\boxed{166}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1991|num-b=2|num-a=4}}</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/August_1,_2011&diff=41085AoPS Wiki talk:Problem of the Day/August 1, 20112011-08-01T12:23:27Z<p>Rbhale12: </p>
<hr />
<div>==Problem==<br />
{{:AoPSWiki:Problem of the Day/August 1, 2011}}<br />
==Solution==<br />
{{potd_solution}}<br />
We can split this summation, as shown:<br />
<math> \sum_{k = 1}^{\infty}{\frac{8+2^{k}}{4^{k}}} = \sum_{k = 1}^{\infty}{\frac{8}{4^{k}}} +\sum_{k = 1}^{\infty}{\frac{2^k}{4^{k}}} </math>.<br />
<br />
Now we must simply find each of the smaller sums, and add them.<br />
<br />
<math>\sum_{k = 1}^{\infty}{\frac{8}{4^{k}}}</math>:<br />
<br />
We can use the formula for the sum of an infinite geometric series (<math>\frac{a}{1-r}</math> where <math>a</math> is the first term and <math>r</math> is the common ratio).<br />
<math>a=\frac{8}{4}=2</math> and <math>r=\frac{1}{4}</math> since each term is getting multiplied by <math>\frac{1}{4}</math> to receive the next term. Therefore, this sum is:<br />
<math>\frac{2}{1-\frac{1}{4}}=\frac{2}{\frac{3}{4}}=\frac{8}{3}</math>.<br />
<br />
<math>\sum_{k = 1}^{\infty}{\frac{2^k}{4^{k}}}</math>:<br />
<br />
Similarly, we can use the formula used to solve the first part.<br />
<math>a=\frac{2}{4}=\frac{1}{2}</math> and <math>r=\frac{1}{2}</math>. Therefore, this sum is: <math>\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1</math>.<br />
<br />
Using these two answers, the desired sum is: <math>\frac{8}{3}+1=\boxed{\frac{11}{3}}</math>.</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=Joining_an_ARML_team&diff=38164Joining an ARML team2011-04-19T20:24:10Z<p>Rbhale12: </p>
<hr />
<div>Team selection for the [[American Regions Mathematics League]] varies from team to team.<br />
<br />
==Arizona==<br />
<br />
The Arizona ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
== Connecticut ==<br />
Connecticut team selection is based on performance in math leagues across the state, from which a few upperclassmen can automatically qualify, and a runoff which is held in early March. To participate in the runoff, either have the math team coach at your school contact the league director in February to get the date or, if you don't participate in a league, contact Daniel Bochicchio, the current coach, at dbochicchio@eosmith.org. Connecticut sends three teams.<br />
<br />
== Florida ==<br />
Florida ARML sends three teams to ARML each year. The selection criteria for the Florida ARML team takes into consideration several factors:<br />
<br />
*[[AMC]] and [[AIME]] performance<br />
*Past [[AMC]], [[AIME]], and [[USAMO]] scores<br />
*Past [[ARML]] performance<br />
*FAMAT-designated competitions<br />
*An annual statewide tryout test <br />
<br />
Florida ARML is organized by the Florida Student Association of Mathematics. Email eliross2@aol.com for more information.<br />
<br />
== Georgia ==<br />
<br />
[[Georgia ARML]] has sent two teams to ARML each year since 1991. In the past, the A team competed in the A division and the B team in the B division, but recently, the B team placed highly in the B division and so will compete in the A division at least until 2009. Usually, three alternates compete as well. Students interested in participation should do well in [[Georgia mathematics competitions | local tournaments]] and the [[AMC]] and [[AIME]]. In addition to schools invited to the annual [[GCTM State Math Tournament | varsity state tournament]], the Georgia ARML coaches invite other individuals that are under serious consideration for the ARML team.The coaches select the team members during the state tournament based on [[AMC | USAMO index]], performance in local tournaments, and score at the state tournament. Underclassmen are given leeway, as they have years to improve: middle schoolers, however, are rarely selected. For more detailed information on team selection, see the [http://paideiaschool.org/TeacherPages/Steve_Sigur/resources/GAARML/GEORGIAARML/Welcome.html Georgia ARML website].<br />
<br />
The team usually practices on Sundays from the state tournament until the trip to ARML. The specific compositions of the A and B teams are not usually determined until immediately before ARML. Whether a team member will be on the A team, will be on the B team, or will be an alternate depends on the person's performance against other team members in practice individual rounds and the coaches's discretion.<br />
<br />
== Maine ==<br />
<br />
The two Maine ARML Teams consist of approximately the top 30 scorers on 5 [[MAML]] (Maine Association of Math Leagues) Meets. Training includes the problem set "Pete's Fabulous 42."<br />
<br />
* [http://www.maml.net MAML Website]<br />
<br />
== Montgomery (Montgomery County, MD) ==<br />
<br />
Montgomery County typically sends four teams of high-schoolers and one team of middle-schoolers to ARML, with Montgomery A competing in division A and the other teams competing in division B. (with the exception of Montgomery B, which will be competing in division A for 2007-2009)<br />
<br />
Top scorers from the Montgomery County high school math league are invited to ARML practices after the regular season, which typically ends around February. Top scorers from the middle school math league are also invited. However, interested students who did not participate in the league or weren't invited are still welcome to join and should contact Eric Walstein at Montgomery Blair High School.<br />
<br />
Practices are usually held at Montgomery Blair High School on Thursdays from 7-9pm. (someone will need to verify this, my info is about 5 years old) Team selection is done by individual scores at practice and at the discretion of the coaches.<br />
<br />
==Minnesota==<br />
<br />
Minnesota sends two teams to ARML each year, with the Gold and Maroon teams usually competing in divisions A and B, respectively.<br />
<br />
Roughly 35 students are invited to ARML practices, which take place on three consecutive Saturdays in May. There is no practice Memorial Day weekend. Invitations to the ARML team are extended to the top 10 state scorers on the AMC12, the top 10 regular-season scorers on the Minnesota High School Math League, and the top 10 scorers on the Invitational Event at the statewide math league tournament (held in March).<br />
<br />
Since these lists tend to overlap quite a bit, invitations are usually given to students "further down" these lists until enough invites have been given to fill two 15-person teams. <br />
<br />
In addition, an extra 5 or so younger students (typically in grades 8 through 10) are invited to be ARML "students in training". They may or may not go to ARML, but often serve as alternates (in the event that other students cannot attend). The expectation is that a student in training will learn from the practices, and the following year will be on one of the two teams. Since the creation of the AMC8 and AMC10 exams, the top scorers from these exams have typically been invited to ARML practices, either as team members or students in training.<br />
<br />
The selection of the Gold and Maroon teams is determined by students' performance at practices, and is not announced until the night before the competition.<br />
<br />
Usually at least one student in training is invited to go to ARML. This is to prevent a last-minute no-show (due to illness or emergency, for example) from crippling one of the teams.<br />
<br />
Invitations to participate on the MN team are usually sent out shortly after the MN State High School Math League statewide tournament in March. <br />
<br />
* [http://www.macalester.edu/mathleague/index.htm Minnesota State High School Math League site]<br />
<br />
==Missouri==<br />
<br />
Missouri sends two teams, Red and Blue, and a few alternates for a total of 36-37 students to the Iowa site ARML competition.<br />
<br />
Open practices are held usually once a month during the school year in two places, Springfield and St. Louis. Attendance is encouraged but not required. There are usually one or two "all day practices" a year at Missouri University of Science and Technology.<br />
<br />
Applications are based on AMC/AIME scores, GPML scores, other math competitions, and past ARML experience. Most if not all applicants are selected for the team.<br />
<br />
Since ARML is NOT a strictly a “state” team competition and because neither Kansas nor southern Illinois have ARML teams, ARML has given the “Missouri” team permission to include students from Kansas and from western Illinois (eg Edwardsville, etc.) on the MO ARML team. (eg. the 2008 team fielded 6 from KS, and 1 from IL)<br />
<br />
[http://math.missouristate.edu/MissouriARML.htm MO ARML homepage]<br />
<br />
==Nevada==<br />
<br />
The Nevada ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==New Jersey==<br />
New Jersey sends three different ARML teams; the three being Central Jersey (which typically sends an A team and a B team), Bergen County, and AAST. <br />
<br />
=== Central Jersey ===<br />
Qualification for the Central Jersey ARML team requires that<br />
*the student has participated (and chosen as a representative for the school) in at least 4 of the Central Jersey Math League meets,<br />
*and the average score meets a certain minimum which varies per year (this year, it is 3.0).<br />
<br />
Afterwards, interested students have to attend training sessions which take place at Highland Park High School or Hillsborough.<br />
<br />
More information can be found at [http://uzza.us/cjml/arml.html CJML website].<br />
<br />
==New Mexico==<br />
<br />
The New Mexico ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==New York City==<br />
For information about the New York City Math Team, please visit [http://www.nycmathteam.com the NYC Math Team homepage].<br />
<br />
==North Carolina==<br />
<br />
North Carolina sends two 15-person ARML teams to compete at Penn State. The 2006 NC "A" team placed 1st in Division A. Invitations to spring team practice sessions are extended based on performances at the [http://courses.ncssm.edu/goebel/statecon/state.htm NC State Math Contest], AMC, AIME, Duke Math Meet and other math competitions. The top 15 or so finishers at the NC State Math Contest (Comprehensive) and all USAMO qualifiers are among those typically invited to practice sessions. Performance at these practice sessions and math contests determine team assignments. <br />
<br />
[http://courses.ncssm.edu/goebel/statecon/Inform/sites.htm NC State Math Contest qualifying events] are held throughout the state in February and March. Archie Benton, John Noland and several others coach and prepare the team with e-mail problems and the spring practice sessions held at the NC School of Science and Mathematics in Durham.<br />
<br />
==Ohio==<br />
<br />
Invitation is based on OCTM (a state-wide competition) and AMC scores. Also, at the OHMIO (second level of OCTM) the offer to join is extended to anyone who is interested. Team placement is based on a combination of OCTM scores, AMC scores, and how well the person does in practices. Ohio normally sends two teams, but is sending three this year because enough students were interested. Also, starting this year, the Ohio A team is competing in division A. The other two teams are competing in division B.<br />
<br />
The first practice is Sunday, April 22, and the second is Saturday, May 19.<br />
<br />
==Pennsylvania==<br />
===Lehigh Valley===<br />
For information about the Lehigh Valley ARML teams, please visit [http://www.lehigh.edu/dmd1/public/www-data/arml.html the Lehigh Valley ARML homepage] and [http://www.lehigh.edu/dmd1/public/www-data/logistics.html the Lehigh Valley ARML logistics page.].<br />
<br />
==San Diego==<br />
<br />
The SD team is entering just its third year of participation, so a permanent process of selecting team members has yet to be decided upon. The team is organized by and practices at the San Diego Math Circle (SDMC), and most of the students on last year's team were regular attendees at SDMC. Also, since the 2007 team contained no seniors, the organizers for the 2008 team are not planning on extending invites to new students unless scores on the AMC exams or the San Diego Math Olympiad (SDMO) are particularly high. <br />
<br />
A student who wishes to attend practice should try to take the SDMO. If a student wishes to speak with one of the coaches for the team, they may do so by contacting AoPS member 'Generating'.<br />
<br />
* [http://www.sdmathcircle.org/Welcome.php San Diego Math Circle website]<br />
<br />
==San Francisco Bay Area==<br />
<br />
The San Francisco Bay Area ARML team has been organized by [[mathleague.org]] since 2008. More information on participating can be found at [http://mathleague.org/arml http://mathleague.org/arml].<br />
<br />
==South Carolina==<br />
<br />
To join the SC All-State Team, one must take a preliminary exam administered through their school. For more information, please contact ronskimomo@hotmail.com.<br />
<br />
The preliminary exam is composed of 25 questions (non multiple choice), and is usually composed of easy to mid range AMC-12 level questions. From this exam, approximately 50-60 (in 2006 it was 49) of the top scorers from the state are selected into the South Carolina All State Mathematics Team. The qualifying floor this year was 11 out of the 25 questions. After an individual is accepted into the SC All State Team, he or she is invited to one or two ARML practices which are usually composed of individual tests, team tests, and a power round test.<br />
<br />
<br />
* [http://scall-statemathteam.com/default.aspx Official SC ARML site]<br />
* [[South Carolina ARML | SC ARML wiki page]]<br />
<br />
== Southern California ==<br />
<br />
The Southern California team is open to residents of the following Southern California counties: Santa Barbara, Ventura, Los Angeles, Orange, Kern, San Bernardino, and Riverside. The organization fields three or four teams (45 or 60 students) and competes at the western ARML site in Las Vegas.<br />
<br />
Practices are held throughout the school year, approximately once a month, on the campus of California State University, Long Beach. Practices are normally held on Saturday afternoons. In addition, there is a Santa Barbara area group that meets and practices in Goleta and becomes part of the Southern California team.<br />
<br />
Team selection is based on all of the following criteria:<br />
<br />
* Attendence at practice sessions.<br />
* Performance on problems at practice sessions.<br />
* Performance at ARML itself in previous years.<br />
* Performance on AMC and AIME, including current and previous years.<br />
* Performance at CSULB Math Day at the Beach, a contest held in March.<br />
<br />
The coach is Dr. Kent Merryfield, a professor at CSULB. His AoPS user name is Kent Merryfield. Please contact him for further information.<br />
<br />
* [http://www.csulb.edu/depts/math/?q=node/20 Math Day at the Beach website.]<br />
* [http://www.csulb.edu/depts/math/?q=node/32 SoCal ARML website.]<br />
* [[Southern California ARML | SoCal ARML wiki page.]]<br />
<br />
==Texas==<br />
<br />
The coach, Sam Baethge chooses mathletes around Texas (or schools in Texas) who have done well in various math contests. The keys are the AMCs, and Mathcounts. TXML also helps, as he runs the competition. The TML, or Texas Math League takes the team to ARML in Iowa. Three teams of 15 students each go. There is a gold, silver, and unofficial team.<br />
<br />
==West Virginia==<br />
<br />
The West Virginia team is selected using the top 15 winners in West Virginia State Math Field Day. Winners 16-30 are used as potential alternates for the team. West Virginia State Math Field Day uses a similar format as the ARML, having an Individual Exam, Individual Short Answer Section, Team Questions, a Team Power Question, and 2 sets of relays of 5 each (there are 10 members in each team).<br />
<br />
==See Other==<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=820893#820893 Newer discussion on AoPS message boards.]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40434 Older discussion on AoPS message boards.]<br />
* [[ARML]]<br />
[[Category:ARML]]</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=370412011 AMC 10A Problems2011-02-21T17:44:00Z<p>Rbhale12: /* Problem 23 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A small bottle of shampoo can hold 35 milliliters of shampoo, whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?<br />
<br />
<math> \textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Suppose [<math>a</math> <math>b</math>] denotes the average of <math>a</math> and <math>b</math>, and {<math>a</math> <math>b</math> <math>c</math>} denotes the average of <math>a</math>, <math>b</math>, and <math>c</math>. What is {{1 1 0} [0 1] 0}?<br />
<br />
<math> \textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Let <math>X</math> and <math>Y</math> be the following sums of arithmetic sequences: <cmath> \begin{eqnarray*} X &=& 10 + 12 + 14 + \cdots + 100, \\ Y &=& 12 + 14 + 16 + \cdots + 102. \end{eqnarray*} </cmath> What is the value of <math>Y - X</math>?<br />
<br />
<math> \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 </math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of <math>12</math>, <math>15</math>, and <math>10</math> minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students? <br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ \frac{37}{3} \qquad\textbf{(C)}\ \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
Set <math>A </math> has 20 elements, and set <math>B </math> has 15 elements. What is the smallest possible number of elements in <math>A \cup B </math>, the union of <math>A </math> and <math>B </math>?<br />
<br />
<math> \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the following equations does NOT have a solution?<br />
<br />
<math>\text{(A)}\:(x+7)^2=0</math><br />
<br />
<math>\text{(B)}\:|-3x|+5=0</math><br />
<br />
<math>\text{(C)}\:\sqrt{-x}-2=0</math><br />
<br />
<math>\text{(D)}\:\sqrt{x}-8=0</math><br />
<br />
<math>\text{(E)}\:|-3x|-4=0 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<br />
<math> \textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A rectangular region is bounded by the graphs of the equations <math>y=a, y=-b, x=-c,</math> and <math>x=d</math>, where <math>a,b,c,</math> and <math>d</math> are all positive numbers. Which of the following represents the area of this region?<br />
<br />
<math> \textbf{(A)}\ ac+ad+bc+bd\qquad\textbf{(B)}\ ac-ad+bc-bd\qquad\textbf{(C)}\ ac+ad-bc-bd \quad\quad\qquad\textbf{(D)}\ -ac-ad+bc+bd\qquad\textbf{(E)}\ ac-ad-bc+bd </math><br />
<br />
[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A majority of the 30 students in Ms. Deameanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was <math>\</math><math>17.71</math>. What was the cost of a pencil in cents?<br />
<br />
<math>\text{(A)}\,7 \qquad\text{(B)}\,11 \qquad\text{(C)}\,17 \qquad\text{(D)}\,23 \qquad\text{(E)}\,77</math><br />
<br />
[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Square <math>EFGH</math> has one vertex on each side of square <math>ABCD</math>. Point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=7\cdot EB</math>. What is the ratio of the area of <math>EFGH</math> to the area of <math>ABCD</math>?<br />
<br />
<math>\text{(A)}\,\frac{49}{64} \qquad\text{(B)}\,\frac{25}{32} \qquad\text{(C)}\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?<br />
<br />
<math>\text{(A)}\,13 \qquad\text{(B)}\,14 \qquad\text{(C)}\,15 \qquad\text{(D)}\,16 \qquad\text{(E)}\,17</math><br />
<br />
<br />
[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?<br />
<br />
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math><br />
<br />
[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?<br />
<br />
<math>\text{(A)}\,\frac{1}{36} \qquad\text{(B)}\,\frac{1}{12} \qquad\text{(C)}\,\frac{1}{6} \qquad\text{(D)}\,\frac{1}{4} \qquad\text{(E)}\,\frac{5}{18}</math><br />
<br />
[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?<br />
<br />
<math>\text{(A)}\,140 \qquad\text{(B)}\,240 \qquad\text{(C)}\,440 \qquad\text{(D)}\,640 \qquad\text{(E)}\,840</math><br />
<br />
[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>?<br />
<br />
<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math><br />
<br />
[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
In the eight-term sequence <math>A,B,C,D,E,F,G,H</math>, the value of <math>C</math> is 5 and the sum of any three consecutive terms is 30. What is <math>A+H</math>?<br />
<br />
<math>\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43</math><br />
<br />
[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?<br />
<br />
<math><br />
\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad<br />
\textbf{(B)}\ \frac{\pi}{2} \qquad<br />
\textbf{(C)}\ 2 \qquad<br />
\textbf{(D)}\ \frac{3\pi}{4} \qquad<br />
\textbf{(E)}\ 1+\frac{\pi}{2}} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?<br />
<br />
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?<br />
<br />
<math> \textbf{(A)}\ \frac{7}{11}\qquad\textbf{(B)}\ \frac{9}{13}\qquad\textbf{(C)}\ \frac{11}{15}\qquad\textbf{(D)}\ \frac{15}{19}\qquad\textbf{(E)}\ \frac{15}{16} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
Seven students count from 1 to 1000 as follows:<br />
<br />
•Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.<br />
<br />
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.<br />
<br />
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.<br />
<br />
•Finally, George says the only number that no one else says.<br />
<br />
What number does George say?<br />
<br />
<math> \textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998 </math><br />
<br />
[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?<br />
<br />
<math> \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} </math><br />
<br />
[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<br />
<br />
<math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math><br />
<br />
[[2011 AMC 10A Problems/Problem 25|Solution]]</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_1&diff=360652008 AIME I Problems/Problem 12010-11-21T01:05:50Z<p>Rbhale12: </p>
<hr />
<div>== Problem ==<br />
Of the students attending a school party, <math>60\%</math> of the students are girls, and <math>40\%</math> of the students like to dance. After these students are joined by <math>20</math> more boy students, all of whom like to dance, the party is now <math>58\%</math> girls. How many students now at the party like to dance?<br />
<br />
== Solution ==<br />
==Solution 1==<br />
Say that there were <math>3k</math> girls and <math>2k</math> boys at the party originally. <math>2k</math> like to dance. Then, there are <math>3k</math> girls and <math>2k + 20</math> boys, and <math>2k + 20</math> like to dance.<br />
<br />
Thus, <math>\dfrac{3k}{5k + 20} = \dfrac{29}{50}</math>, solving gives <math>k = 116</math>. Thus, the number of people that like to dance is <math>2k + 20 = \boxed{252}</math>.<br />
<br />
==Solution 2==<br />
Let the number of girls be <math>g</math>. Let the number of total people originally be <math>t</math>.<br />
<br />
We know that <math>\frac{g}{t}=\frac{3}{5}</math> from the problem.<br />
<br />
We also know that <math>\frac{g}{t+20}=\frac{29}{50}</math> from the problem.<br />
<br />
We now have a system and we can solve.<br />
<br />
The first equation becomes:<br />
<br />
<math>3t=5g</math>.<br />
<br />
The second equation becomes: <br />
<br />
<math>50g=29t+580</math><br />
<br />
Now we can sub in <math>30t=50g</math> by multiplying the first equation by <math>10</math>. We can plug this into our second equation.<br />
<br />
<math>30t=29t+580</math><br />
<br />
<math>t=580</math><br />
<br />
We know that there were originally <math>580</math> people. Of those, <math>\frac{2}{5}*580=232</math> like to dance.<br />
<br />
We also know that with these people, <math>20</math> boys joined, all of whom like to dance. We just simply need to add <math>20</math> to get <math>232+20=\boxed{252}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=I|before=First Question|num-a=2}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_20&diff=360312007 AMC 10A Problems/Problem 202010-11-12T15:05:30Z<p>Rbhale12: </p>
<hr />
<div>== Problem ==<br />
Suppose that the number <math>a</math> satisfies the equation <math>4 = a + a^{ - 1}</math>. What is the value of <math>a^{4} + a^{ - 4}</math>?<br />
<br />
<math>\text{(A)}\ 164 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 192 \qquad \text{(D)}\ 194 \qquad \text{(E)}\ 212</math><br />
<br />
__NOTOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Notice that <math>(a^{k} + a^{-k})^2 = a^{2k} + a^{-2k} + 2</math>. Thus <math>a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}</math>.<br />
<br />
=== Solution 2 ===<br />
<math>4a = a^2 + 1</math>. We apply the [[quadratic formula]] to get <math>a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}</math>. <br />
<br />
Thus <math>a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4</math> (so it doesn't matter which root of <math>a</math> we use). Using the [[binomial theorem]] we can expand this out and collect terms to get <math>194</math>. <br />
<br />
=== Solution 3 ===<br />
<br />
We know that <math>a+\frac{1}{a}=4</math>. We can square both sides to get <math>a^2+\frac{1}{a^2}+2=16</math>, so <math>a^2+\frac{1}{a^2}=14</math>. Squaring both sides again gives <math>a^4+\frac{1}{a^4}+2=14^2=196</math>, so <math>a^4+\frac{1}{a^4}=\boxed{194}</math>. Q. E. D.<br />
== See also ==<br />
{{AMC10 box|year=2007|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_1&diff=358182006 AIME II Problems/Problem 12010-09-19T17:47:15Z<p>Rbhale12: /* Solution */</p>
<hr />
<div>== Problem ==<br />
In [[convex polygon|convex]] [[hexagon]] <math>ABCDEF</math>, all six sides are congruent, <math>\angle A</math> and <math>\angle D</math> are [[right angle]]s, and <math>\angle B, \angle C, \angle E,</math> and <math>\angle F</math> are [[congruent]]. The area of the hexagonal region is <math>2116(\sqrt{2}+1).</math> Find <math>AB</math>.<br />
<br />
== Solution ==<br />
Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>. <br />
<br />
[[Image:2006_II_AIME-1.png]]<br />
<br />
The diagonal <math>BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}</math>. Then the areas of the triangles AFB and CDE in total are <math>\frac{x^2}{2}\cdot 2</math>,<br />
and the area of the rectangle BCEF equals <math>x\cdot x\sqrt{2}=x^2\sqrt{2}</math><br />
<br />
Then we have to solve the equation <br />
<div style="text-align:center;"><br />
<math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>.<br />
<br />
<math>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</math><br />
<br />
<math>2116=x^2</math><br />
<br />
<math>x=46</math></div><br />
<br />
Therefore, <math>AB</math> is <math>\boxed{046}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2006|n=II|before=First Question|num-a=2}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems&diff=356921993 AIME Problems2010-08-24T19:58:47Z<p>Rbhale12: /* Problem 2 */</p>
<hr />
<div>{{AIME Problems|year=1993}}<br />
<br />
== Problem 1 ==<br />
How many even integers between 4000 and 7000 have four different digits? <br />
<br />
[[1993 AIME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went <math>\frac{n^{2}}{2}</math> miles on the <math>n^{\mbox{th}}_{}</math> day of this tour, how many miles was he from his starting point at the end of the <math>40^{\mbox{th}}_{}</math> day? <br />
<br />
[[1993 AIME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught <math>n\,</math> fish for various values of <math>n\,</math>. <br />
<center><math>\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\<br />
\hline \text{number of contestants who caught} \ n \ \text{fish} & 9 & 5 & 7 & 23 & \dots & 5 & 2 & 1 \\<br />
\hline \end{array}</math></center><br />
In the newspaper story covering the event, it was reported that <br />
:(a) the winner caught <math>15</math> fish; <br />
:(b) those who caught <math>3</math> or more fish averaged <math>6</math> fish each; <br />
:(c) those who caught <math>12</math> or fewer fish averaged <math>5</math> fish each. <br />
What was the total number of fish caught during the festival?<br />
<br />
[[1993 AIME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
How many ordered four-tuples of integers <math>(a,b,c,d)\,</math> with <math>0 < a < b < c < d < 500\,</math> satisfy <math>a + d = b + c\,</math> and <math>bc - ad = 93\,</math>?<br />
<br />
[[1993 AIME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Let <math>P_0(x) = x^3 + 313x^2 - 77x - 8\,</math>. For integers <math>n \ge 1\,</math>, define <math>P_n(x) = P_{n - 1}(x - n)\,</math>. What is the coefficient of <math>x\,</math> in <math>P_{20}(x)\,</math>?<br />
<br />
[[1993 AIME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?<br />
<br />
[[1993 AIME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Three numbers, <math>a_1\,</math>, <math>a_2\,</math>, <math>a_3\,</math>, are drawn randomly and without replacement from the set <math>\{1, 2, 3, \dots, 1000\}\,</math>. Three other numbers, <math>b_1\,</math>, <math>b_2\,</math>, <math>b_3\,</math>, are then drawn randomly and without replacement from the remaining set of 997 numbers. Let <math>p\,</math> be the probability that, after a suitable rotation, a brick of dimensions <math>a_1 \times a_2 \times a_3\,</math> can be enclosed in a box of dimensions <math>b_1 \times b_2 \times b_3\,</math>, with the sides of the brick parallel to the sides of the box. If <math>p\,</math> is written as a fraction in lowest terms, what is the sum of the numerator and denominator?<br />
<br />
[[1993 AIME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Let <math>S\,</math> be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of <math>S\,</math> so that the union of the two subsets is <math>S\,</math>? The order of selection does not matter; for example, the pair of subsets <math>\{a, c\}\,</math>, <math>\{b, c, d, e, f\}\,</math> represents the same selection as the pair <math>\{b, c, d, e, f\}\,</math>, <math>\{a, c\}\,</math>.<br />
<br />
[[1993 AIME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Two thousand points are given on a circle. Label one of the points 1. From this point, count 2 points in the clockwise direction and label this point 2. From the point labeled 2, count 3 points in the clockwise direction and label this point 3. (See figure.) Continue this process until the labels <math>1,2,3\dots,1993\,</math> are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as 1993? <br />
<br />
[[Image:AIME_1993_Problem_9.png]]<br />
<br />
[[1993 AIME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Euler's formula states that for a convex polyhedron with <math>V\,</math> vertices, <math>E\,</math> edges, and <math>F\,</math> faces, <math>V-E+F=2\,</math>. A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon. At each of its <math>V\,</math> vertices, <math>T\,</math> triangular faces and <math>P^{}_{}</math> pentagonal faces meet. What is the value of <math>100P+10+V\,</math>? <br />
<br />
[[1993 AIME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is <math>m/n\,</math>, where <math>m\,</math> and <math>n\,</math> are relatively prime positive integers. What are the last three digits of <math>m+n\,</math>? <br />
<br />
[[1993 AIME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The vertices of <math>\triangle ABC</math> are <math>A = (0,0)\,</math>, <math>B = (0,420)\,</math>, and <math>C = (560,0)\,</math>. The six faces of a die are labeled with two <math>A\,</math>'s, two <math>B\,</math>'s, and two <math>C\,</math>'s. Point <math>P_1 = (k,m)\,</math> is chosen in the interior of <math>\triangle ABC</math>, and points <math>P_2\,</math>, <math>P_3\,</math>, <math>P_4, \dots</math> are generated by rolling the die repeatedly and applying the rule: If the die shows label <math>L\,</math>, where <math>L \in \{A, B, C\}</math>, and <math>P_n\,</math> is the most recently obtained point, then <math>P_{n + 1}^{}</math> is the midpoint of <math>\overline{P_n L}</math>. Given that <math>P_7 = (14,92)\,</math>, what is <math>k + m\,</math>?<br />
<br />
[[1993 AIME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let <math>t\,</math> be the amount of time, in seconds, before Jenny and Kenny can see each other again. If <math>t\,</math> is written as a fraction in lowest terms, what is the sum of the numerator and denominator?<br />
<br />
[[1993 AIME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called ''unstuck'' if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form <math>\sqrt{N}\,</math>, for a positive integer <math>N\,</math>. Find <math>N\,</math>.<br />
<br />
[[1993 AIME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Let <math>\overline{CH}</math> be an altitude of <math>\triangle ABC</math>. Let <math>R\,</math> and <math>S\,</math> be the points where the circles inscribed in the triangles <math>ACH\,</math> and <math>BCH^{}_{}</math> are tangent to <math>\overline{CH}</math>. If <math>AB = 1995\,</math>, <math>AC = 1994\,</math>, and <math>BC = 1993\,</math>, then <math>RS\,</math> can be expressed as <math>m/n\,</math>, where <math>m\,</math> and <math>n\,</math> are relatively prime integers. Find <math>m + n\,</math>.<br />
<br />
[[1993 AIME Problems/Problem 15|Solution]]<br />
<br />
== See also ==<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:AIME Problems]]</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_11&diff=356562010 AMC 10A Problems/Problem 112010-08-16T22:14:29Z<p>Rbhale12: </p>
<hr />
<div>Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>. <br />
<br />
<math> a\le 2x+3\le b </math><br />
<br />
Subtract <math> 3 </math> from all of the quantities:<br />
<br />
<math> a-3\le 2x\le b-3 </math><br />
<br />
Divide all of the quantities by <math> 2 </math>.<br />
<br />
<math> \frac{a-3}{2}\le x\le \frac{b-3}{2} </math><br />
<br />
Since we have the range of the solutions, we can make them equal to <math> 10 </math>.<br />
<br />
<math> \frac{b-3}{2}-\frac{a-3}{2} = 10 </math><br />
<br />
Multiply both sides by 2.<br />
<br />
<math> (b-3) - (a-3) = 20 </math><br />
<br />
Re-write without using parentheses.<br />
<br />
<math> b-3-a+3 = 20 </math><br />
<br />
Simplify.<br />
<br />
<math> b-a = 20 </math><br />
<br />
We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_11&diff=356552010 AMC 10A Problems/Problem 112010-08-16T22:12:25Z<p>Rbhale12: </p>
<hr />
<div>Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>. <br />
<br />
<math> a\le 2x+3\le b </math><br />
<br />
Subtract <math> 3 </math> from all of the quantities:<br />
<br />
<math> a-3\le 2x\le b-3 </math><br />
<br />
Divide all of the quantities by <math> 2 </math>.<br />
<br />
<math> \frac{a-3}{2}\le x\le \frac{b-3}{2} </math><br />
<br />
Since we have the range of the solutions, we can make them equal to <math> 10 </math>.<br />
<br />
<math> \frac{b-3}{2}-\frac{a-3}{2} = 10 </math><br />
<br />
Multiply both sides by 2.<br />
<br />
<math> (b-3) - (a-3) = 20 </math><br />
<br />
Simplify<br />
<br />
<math> b-3-a+3 = 20 \RightArrow b-a = 20 </math><br />
<br />
We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_11&diff=356542010 AMC 10A Problems/Problem 112010-08-16T22:09:50Z<p>Rbhale12: Created page with 'Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>. <math> a\le…'</p>
<hr />
<div>Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>. <br />
<br />
<math> a\le 2x+3\le b </math><br />
<br />
Subtract <math> 3 </math> from all of the quantities:<br />
<br />
<math> a-3\le 2x\le b-3 </math><br />
<br />
Divide all of the quantities by <math> 2 </math>.<br />
<br />
<math> \frac{a-3}{2}\le x\le \frac{b-3}{2} </math><br />
<br />
Since we have the range of the solutions, we can make them equal to <math> 99 </math>.<br />
<br />
<math> \frac{b-3}{2}-\frac{a-3}{2} = 10 </math><br />
<br />
Multiply both sides by 2.<br />
<br />
<math> (b-3) - (a-3) = 20 </math><br />
<br />
Simplify<br />
<br />
<math> b-3-a+3 = 20 \RightArrow b-a = 20 </math><br />
<br />
We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_14&diff=356002010 AMC 10B Problems/Problem 142010-08-09T04:11:09Z<p>Rbhale12: </p>
<hr />
<div>We must find the average of the numbers from <math> 1 </math> to <math> 99 </math> and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives:<br />
<br />
100(100x)=99(50)+x,<br />
10000x=99(50)+x,<br />
9999x=99(50),<br />
101x=50,<br />
<math><br />
x=\frac{50}{101}<br />
</math><br />
This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_14&diff=355992010 AMC 10B Problems/Problem 142010-08-09T04:08:40Z<p>Rbhale12: </p>
<hr />
<div>We must find the average of the numbers from <math> 1 </math> to <math> 99 </math> and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives:<br />
<math><br />
100(100x)=99(50)+x,<br />
10000x=99(50)+x,<br />
9999x=99(50),<br />
101x=50,<br />
x=\frac{50}{101}.<br />
</math><br />
This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_14&diff=355982010 AMC 10B Problems/Problem 142010-08-09T04:06:26Z<p>Rbhale12: </p>
<hr />
<div>We must find the average of the numbers from <math> 1 </math> to <math> 99 </math> and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives:<br />
<math><br />
\linebreak<br />
100(100x)=99(50)+x <br />
\linebreak<br />
10000x=99(50)+x <br />
\linebreak<br />
9999x=99(50) <br />
\linebreak<br />
101x=50 <br />
\linebreak<br />
x=\frac{50}{101}<br />
\linebreak<br />
</math><br />
This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_14&diff=355972010 AMC 10B Problems/Problem 142010-08-09T04:02:48Z<p>Rbhale12: </p>
<hr />
<div>We must find the average of the numbers from <math> 1 </math> to <math> 99 </math> and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives:<br />
<math><br />
100(100x)=99(50)+x <br />
10000x=99(50)+x <br />
9999x=99(50) <br />
101x=50 <br />
x=\frac{50}{101}<br />
</math><br />
This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_14&diff=355962010 AMC 10B Problems/Problem 142010-08-09T04:01:21Z<p>Rbhale12: </p>
<hr />
<div>We must find the average of the numbers from 1 to 99 and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives:<br />
<math><br />
100(100x)=99(50)+x<br />
10000x=99(50)+x<br />
9999x=99(50)<br />
101x=50<br />
x=\frac{50}{101}<br />
</math><br />
This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_14&diff=355952010 AMC 10B Problems/Problem 142010-08-09T03:58:17Z<p>Rbhale12: </p>
<hr />
<div>We must find the average of the numbers from 1 to 99 and x in terms of x. The sum of all these terms is \frac{99(100)}{2}+x=99(50)+x. We must divide this by the total number of terms, which is 100. We get: \frac{99(50)+x}{100}. This is equal to 100x, as stated in the problem. We have: \frac{99(50)+x}{100}=100x. We can now cross multiply. This gives:<br />
<br />
100(100x)=99(50)+x<br />
10000x=99(50)+x<br />
9999x=99(50)<br />
101x=50<br />
x=\frac{50}{101}<br />
<br />
This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_14&diff=355942010 AMC 10B Problems/Problem 142010-08-09T03:56:17Z<p>Rbhale12: Created page with 'We must find the average of the numbers from 1 to 99 and x in terms of x. The sum of all these terms is 99(100)/2+x=99(50)+x. We must divide this by the total number of terms, wh…'</p>
<hr />
<div>We must find the average of the numbers from 1 to 99 and x in terms of x. The sum of all these terms is 99(100)/2+x=99(50)+x. We must divide this by the total number of terms, which is 100. We get: (99(50)+x)/100. This is equal to 100x, as stated in the problem. We have: (99(50)+x)/100=100x. We can now cross multiply. This gives:<br />
<br />
100(100x)=99(50)+x<br />
10000x=99(50)+x<br />
9999x=99(50)<br />
101x=50<br />
x=50/101<br />
<br />
This gives us our answer. <math> \boxed{\mathrm{(B)}= 50/101} </math></div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems&diff=355632005 AIME I Problems2010-08-06T13:30:50Z<p>Rbhale12: /* Problem 1 */</p>
<hr />
<div>{{AIME Problems|year=2005|n=I}}<br />
<br />
== Problem 1 ==<br />
Six circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle <math> C </math> with radius 30. Let <math> K </math> be the area of the region inside circle <math> C </math> and outside of the six circles in the ring. Find <math> \lfloor K \rfloor. </math><br />
<br />
[[2005 AIME I Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For each positive integer <math> k, </math> let <math> S_k </math> denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is <math> k. </math> For example, <math> S_3 </math> is the sequence <math> 1,4,7,10,\ldots. </math> For how many values of <math> k </math> does <math> S_k </math> contain the term 2005?<br />
<br />
[[2005 AIME I Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
How many positive integers have exactly three proper divisors, each of which is less than 50?<br />
<br />
[[2005 AIME I Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.<br />
<br />
[[2005 AIME I Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distunguishable arrangements of the 8 coins.<br />
<br />
[[2005 AIME I Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
Let <math> P </math> be the product of the nonreal roots of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math><br />
<br />
[[2005 AIME I Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
In quadrilateral <math> ABCD, BC=8, CD=12, AD=10, </math> and <math> m\angle A= m\angle B = 60^\circ. </math> Given that <math> AB = p + \sqrt{q}, </math> where <math> p </math> and <math> q </math> are positive integers, find <math> p+q. </math><br />
<br />
[[2005 AIME I Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
The equation <math> 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 </math> has three real roots. Given that their sum is <math> \frac mn </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math><br />
<br />
[[2005 AIME I Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge. The 27 cubes are randomly arranged to form a <math> 3\times 3 \times 3 </math> cube. Given the probability of the entire surface area of the larger cube is orange is <math> \frac{p^a}{q^br^c}, </math> where <math> p,q, </math> and <math> r </math> are distinct primes and <math> a,b, </math> and <math> c </math> are positive integers, find <math> a+b+c+p+q+r. </math><br />
<br />
[[2005 AIME I Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Triangle <math> ABC </math> lies in the Cartesian Plane and has an area of 70. The coordinates of <math> B </math> and <math> C </math> are <math> (12,19) </math> and <math> (23,20), </math> respectively, and the coordinates of <math> A </math> are <math> (p,q). </math> The line containing the median to side <math> BC </math> has slope <math> -5. </math> Find the largest possible value of <math> p+q. </math><br />
<br />
[[2005 AIME I Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
A semicircle with diameter <math> d </math> is contained in a square whose sides have length 8. Given the maximum value of <math> d </math> is <math> m - \sqrt{n},</math> find <math> m+n. </math><br />
<br />
[[2005 AIME I Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
For positive integers <math> n, </math> let <math> \tau (n) </math> denote the number of positive integer divisors of <math> n, </math> including 1 and <math> n. </math> For example, <math> \tau (1)=1 </math> and <math> \tau(6) =4. </math> Define <math> S(n) </math> by <math> S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n). </math> Let <math> a </math> denote the number of positive integers <math> n \leq 2005 </math> with <math> S(n) </math> odd, and let <math> b </math> denote the number of positive integers <math> n \leq 2005 </math> with <math> S(n) </math> even. Find <math> |a-b|. </math><br />
<br />
[[2005 AIME I Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
A particle moves in the Cartesian Plane according to the following rules:<br />
<br />
# From any lattice point <math> (a,b), </math> the particle may only move to <math> (a+1,b), (a,b+1), </math> or <math>(a+1,b+1). </math><br />
# There are no right angle turns in the particle's path.<br />
<br />
How many different paths can the particle take from <math> (0,0) </math> to <math> (5,5) </math>?<br />
<br />
[[2005 AIME I Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Consider the points <math> A(0,12), B(10,9), C(8,0), </math> and <math> D(-4,7). </math> There is a unique square <math> S </math> such that each of the four points is on a different side of <math> S. </math> Let <math> K </math> be the area of <math> S. </math> Find the remainder when <math> 10K </math> is divided by 1000.<br />
<br />
[[2005 AIME I Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Triangle <math> ABC </math> has <math> BC=20. </math> The incircle of the triangle evenly trisects the median <math> AD. </math> If the area of the triangle is <math> m \sqrt{n} </math> where <math> m </math> and <math> n </math> are integers and <math> n </math> is not divisible by the square of a prime, find <math> m+n. </math><br />
<br />
[[2005 AIME I Problems/Problem 15|Solution]]<br />
<br />
== See Also ==<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=50 2005 AIME I Math Jam Transcript]<br />
* [[Mathematics competition resources]]</div>Rbhale12https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems&diff=344172006 AMC 10B Problems2010-04-28T22:37:19Z<p>Rbhale12: /* Problem 9 */</p>
<hr />
<div>== Problem 1 ==<br />
What is <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} </math> ?<br />
<br />
<math> \mathrm{(A) \ } -2006\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } 2006 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For real numbers <math>x</math> and <math>y</math>, define <math> x \mathop{\spadesuit} y = (x+y)(x-y) </math>. What is <math> 3 \mathop{\spadesuit} (4 \mathop{\spadesuit} 5) </math>?<br />
<br />
<math> \mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score? <br />
<br />
<math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 17\qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 24 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Circles of diameter 1 inch and 3 inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area? <br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
A <math> 2 \times 3 </math> rectangle and a <math> 3 \times 4 </math> rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square? <br />
<br />
<math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 49\qquad \mathrm{(E) \ } 64 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
A region is bounded by semicircular arcs constructed on the side of a square whose sides measure <math> \frac{2}{\pi} </math>, as shown. What is the perimeter of this region? <br />
<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(.8);<br />
<br />
filldraw( circle( (0,1), 1 ), lightgray, black );<br />
filldraw( circle( (0,-1), 1 ), lightgray, black );<br />
filldraw( circle( (1,0), 1 ), lightgray, black );<br />
filldraw( circle( (-1,0), 1 ), lightgray, black );<br />
filldraw( (-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle, lightgray, black );<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{4}{\pi}\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } \frac{8}{\pi}\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } \frac{16}{\pi} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Which of the folowing is equivalent to <math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} </math> when <math> x < 0 </math> <br />
<br />
<math> \mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle? <br />
<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(.8);<br />
<br />
draw( (-sqrt(5),0) -- (sqrt(5),0), dashed );<br />
draw( (-1,0)--(-1,2)--(1,2)--(1,0)--cycle );<br />
draw( arc( (0,0), sqrt(5), 0, 180 ) );<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 20\pi\qquad \mathrm{(B) \ } 25\pi\qquad \mathrm{(C) \ } 30\pi\qquad \mathrm{(D) \ } 40\pi\qquad \mathrm{(E) \ } 50\pi </math><br />
<br />
[[2006 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade? <br />
<br />
<math> \mathrm{(A) \ } 129\qquad \mathrm{(B) \ } 137\qquad \mathrm{(C) \ } 174\qquad \mathrm{(D) \ } 233\qquad \mathrm{(E) \ } 411 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? <br />
<br />
<math> \mathrm{(A) \ } 43\qquad \mathrm{(B) \ } 44\qquad \mathrm{(C) \ } 45\qquad \mathrm{(D) \ } 46\qquad \mathrm{(E) \ } 47 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math><br />
<br />
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 9 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The lines <math> x=\frac{1}{4}y+a </math> and <math> y=\frac{1}{4}x+b </math> intersect at the point <math> (1,2) </math>. What is <math> a+b </math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } \frac{9}{4} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Joe and JoAnn each bought 12 ounces of coffee in a 16 ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee? <br />
<br />
<math> \mathrm{(A) \ } \frac{6}{7}\qquad \mathrm{(B) \ } \frac{13}{14}\qquad \mathrm{(C) \ }1 \qquad \mathrm{(D) \ } \frac{14}{13}\qquad \mathrm{(E) \ } \frac{7}{6} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
Let <math>a</math> and <math>b</math> be the roots of the equation <math> x^2-mx+2=0 </math>. Suppose that <math> a+(1/b) </math> and <math> b+(1/a) </math> are the roots of the equation <math> x^2-px+q=0 </math>. What is <math>q</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{5}{2}\qquad \mathrm{(B) \ } \frac{7}{2}\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } \frac{9}{2}\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Rhombus <math>ABCD</math> is similar to rhombus <math>BFDE</math>. The area of rhombus <math>ABCD</math> is <math>24</math> and <math> \angle BAD = 60^\circ </math>. What is the area of rhombus <math>BFDE</math>? <br />
<br />
<asy><br />
unitsize(3cm);<br />
defaultpen(.8);<br />
<br />
pair A=(0,0), B=(1,0), D=dir(60), C=B+D;<br />
<br />
draw(A--B--C--D--cycle);<br />
pair Ep = intersectionpoint( B -- (B+10*dir(150)), D -- (D+10*dir(270)) );<br />
pair F = intersectionpoint( B -- (B+10*dir(90)), D -- (D+10*dir(330)) );<br />
<br />
draw(B--Ep--D--F--cycle);<br />
<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",Ep,SW);<br />
label("$F$",F,NE);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 4\sqrt{3}\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 6\sqrt{3} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
Leap Day, February 29, 2004, occured on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur? <br />
<br />
<math> \mathrm{(A) \ } \textrm{Tuesday} \qquad \mathrm{(B) \ } \textrm{Wednesday} \qquad \mathrm{(C) \ } \textrm{Thursday} \qquad \mathrm{(D) \ } \textrm{Friday} \qquad \mathrm{(E) \ } \textrm{Saturday} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same? <br />
<br />
<math> \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{5}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Let <math> a_1 , a_2 , ... </math> be a sequence for which<br />
<br />
<math> a_1=2 </math> , <math> a_2=3 </math>, and <math>a_n=\frac{a_{n-1}}{a_{n-2}} </math> for each positive integer <math> n \ge 3 </math>. <br />
<br />
What is <math> a_{2006} </math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
A circle of radius <math>2</math> is centered at <math>O</math>. Square <math>OABC</math> has side length <math>1</math>. Sides <math>AB</math> and <math>CB</math> are extended past <math>B</math> to meet the circle at <math>D</math> and <math>E</math>, respectively. What is the area of the shaded region in the figure, which is bounded by <math>BD</math>, <math>BE</math>, and the minor arc connecting <math>D</math> and <math>E</math>?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(.8);<br />
<br />
draw( circle( (0,0), 2 ) );<br />
draw( (-2,0) -- (2,0) );<br />
draw( (0,-2) -- (0,2) );<br />
<br />
pair D = intersectionpoint( circle( (0,0), 2 ), (1,0) -- (1,2) );<br />
pair Ep = intersectionpoint( circle( (0,0), 2 ), (0,1) -- (2,1) );<br />
draw( (1,0) -- D );<br />
draw( (0,1) -- Ep );<br />
<br />
filldraw( (1,1) -- arc( (0,0),Ep,D ) -- cycle, mediumgray, black );<br />
<br />
label("$O$",(0,0),SW);<br />
label("$A$",(1,0),S);<br />
label("$C$",(0,1),W);<br />
label("$B$",(1,1),SW);<br />
label("$D$",D,N);<br />
label("$E$",Ep,E);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3})\qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
In rectangle <math>ABCD</math>, we have <math>A=(6,-22)</math>, <math>B=(2006,178)</math>, <math>D=(8,y)</math>, for some integer <math>y</math>. What is the area of rectangle <math>ABCD</math>?<br />
<br />
<math> \mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
For a particular peculiar pair of dice, the probabilities of rolling <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math>, on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the two dice? <br />
<br />
<math> \mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Elmo makes <math>N</math> sandwiches for a fundraiser. For each sandwich he uses <math>B</math> globs of peanut butter at <math>4</math>¢ per glob and <math>J</math> blobs of jam at <math>5</math>¢ per blob. The cost of the peanut butter and jam to make all the sandwiches is <dollar/><math>2.53</math>. Assume that <math>B</math>, <math>J</math>, and <math>N</math> are positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches?<br />
<br />
<math> \mathrm{(A) \ } 1.05\qquad \mathrm{(B) \ } 1.25\qquad \mathrm{(C) \ } 1.45\qquad \mathrm{(D) \ } 1.65\qquad \mathrm{(E) \ } 1.85 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(.8);<br />
<br />
pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br />
pair F = intersectionpoint( A--D, B--Ep );<br />
<br />
draw( A -- B -- C -- cycle );<br />
draw( A -- D );<br />
draw( B -- Ep );<br />
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br />
<br />
label("$7$",(1.25,0.2));<br />
label("$7$",(2.2,0.45));<br />
label("$3$",(0.45,0.35));<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
Circles with centers <math>O</math> and <math>P</math> have radii <math>2</math> and <math>4</math>, respectively, and are externally tangent. Points <math>A</math> and <math>B</math> on the circle with center <math>O</math> and points <math>C</math> and <math>D</math> on the circle with center <math>P</math> are such that <math>AD</math> and <math>BC</math> are common external tangents to the circles. What is the area of the concave hexagon <math>AOBCPD</math>?<br />
<br />
<asy><br />
unitsize(.7cm);<br />
defaultpen(.8);<br />
<br />
pair O = (0,0), P = (6,0), Q = (-6,0);<br />
pair A = intersectionpoint( arc( (-3,0), (0,0), (-6,0) ), circle( O, 2 ) );<br />
pair B = (A.x, -A.y );<br />
pair D = Q + 2*(A-Q);<br />
pair C = Q + 2*(B-Q);<br />
<br />
draw( circle(O,2) );<br />
draw( circle(P,4) );<br />
draw( (Q + 0.8*(A-Q)) -- ( Q + 2.3*(A-Q) ) );<br />
draw( (Q + 0.8*(B-Q)) -- ( Q + 2.3*(B-Q) ) );<br />
draw( A -- O -- B );<br />
draw( C -- P -- D );<br />
draw( O -- P );<br />
<br />
label("$O$",O,W);<br />
label("$P$",P,E);<br />
<br />
label("$A$",A,NNW);<br />
label("$B$",B,SSW);<br />
<br />
label("$D$",D,NNW);<br />
label("$C$",C,SSW);<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2} </math><br />
<br />
[[2006 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is <b><i>not</i></b> the age of one of Mr. Jones's children? <br />
<br />
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2006 AMC 10B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2006 AMC 10B]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=143 2006 AMC B Math Jam Transcript]<br />
* [[Mathematics competition resources]]</div>Rbhale12