https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Bob+Smith&feedformat=atom
AoPS Wiki - User contributions [en]
2024-03-28T08:54:27Z
User contributions
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=92185
AMC historical results
2018-02-21T22:01:11Z
<p>Bob Smith: /* AMC 12A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
<br />
===AMC 10A===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
===AMC 10B===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
===AMC 12A===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
===AMC 12B===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.56<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 111<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 111<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 93<br />
*DHR: 111<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100.5<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100.5<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=92184
AMC historical results
2018-02-21T22:00:39Z
<p>Bob Smith: /* AMC 12A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
<br />
===AMC 10A===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
===AMC 10B===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
===AMC 12A===<br />
*Average score: 151.5<br />
*AIME floor: 10.5<br />
*DHR: 3<br />
<br />
===AMC 12B===<br />
*Average score: TBA<br />
*AIME floor: TBA<br />
*DHR: TBA<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.56<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 111<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 111<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 93<br />
*DHR: 111<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100.5<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100.5<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=92007
AMC historical results
2018-02-19T05:43:45Z
<p>Bob Smith: /* AMC 12A */</p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2018==<br />
===AMC 10A===<br />
*Average Score: To be released on Feb. 20<br />
*AIME floor: To be released on Feb. 20<br />
*Distinguished Honor Roll: To be released on Feb. 20<br />
<br />
===AMC 10B===<br />
*Average score: To be released on Feb. 28<br />
*AIME floor: To be released on Feb. 28<br />
*Distinguished Honor Roll: To be released on Feb. 28<br />
<br />
===AMC 12A===<br />
*Average score: To be released on Feb. 20<br />
*AIME floor: To be released on Feb 20<br />
*Distinguished Honor Roll: To be released on Feb. 20<br />
<br />
===AMC 12B===<br />
*Average score: To be released on Feb. 28<br />
*AIME floor: To be released on Feb. 28<br />
*Distinguished Honor Roll: To be released on Feb. 28<br />
<br />
==2017==<br />
===AMC 10A===<br />
*Average score: 59.33<br />
*AIME floor: 112.5<br />
*DHR: 127.5<br />
<br />
===AMC 10B===<br />
*Average score: 66.56<br />
*AIME floor: 120<br />
*DHR: 136.5<br />
<br />
===AMC 12A===<br />
*Average score: 60.32<br />
*AIME floor: 96<br />
*DHR: 115.5<br />
<br />
===AMC 12B===<br />
*Average score: 58.35<br />
*AIME floor: 100.5<br />
*DHR: 129<br />
<br />
===AIME I===<br />
*Average score: 5.69<br />
*Median score: 6<br />
*USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br />
*USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br />
<br />
===AIME II===<br />
*Average score: 5.64<br />
*Median score: 6<br />
*USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br />
*USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: 65.3<br />
*AIME floor: 110<br />
*DHR: 120<br />
<br />
===AMC 10B===<br />
*Average score: 65.4<br />
*AIME floor: 110<br />
*DHR: 124.5<br />
<br />
===AMC 12A===<br />
*Average score: 59.06<br />
*AIME floor: 92<br />
*DHR: 110<br />
<br />
===AMC 12B===<br />
*Average score: 67.96<br />
*AIME floor: 100<br />
*DHR: 127.5<br />
<br />
===AIME I===<br />
*Average score: 5.83<br />
*Median score: 6<br />
*USAMO cutoff: 220<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 4.33<br />
*Median score: 4<br />
*USAMO cutoff: 205<br />
*USAJMO cutoff: 200<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
*DHR: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
*DHR: 117<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
*DHR: 126<br />
<br />
===AIME I===<br />
*Average score: 5.29<br />
*Median score: 5<br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: 6.63<br />
*Median score: 6<br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
*DHR: 132<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
*DHR: 109.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
*DHR: 121.5<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: 5<br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
*DHR: 117<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
*DHR: 129<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor: 88.5<br />
*DHR: 106.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
*DHR: 108<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: 4<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: 6<br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: 100 (Top 5% (1.00))<br />
<br />
===AIME I===<br />
*Average score: 4.17<br />
*Median score: 4<br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score: 3.27<br />
*Median score: 3<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
none<br />
<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
<br />
==1959==</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_11&diff=67666
2015 AMC 12A Problems/Problem 11
2015-02-06T05:35:56Z
<p>Bob Smith: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
On a sheet of paper, Isabella draws a circle of radius <math>2</math>, a circle of radius <math>3</math>, and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly <math>k \ge 0</math> lines. How many different values of <math>k</math> are possible?<br />
<br />
<math> \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}}\ 5\qquad\textbf{(E)}\ 6</math><br />
<br />
==Solution==<br />
<br />
Isabella can get <math>0</math> lines if the circles are concentric, <math>1</math> if internally tangent, <math>2</math> if overlapping, <math>3</math> if externally tangent, and <math>4</math> if non-overlapping and not externally tangent. There are <math>\boxed{\textbf{(D)}\ 5}</math> values of <math>k</math>.<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2015|ab=A|num-b=10|num-a=12}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_14&diff=67665
2015 AMC 10A Problems/Problem 14
2015-02-06T04:54:57Z
<p>Bob Smith: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externally tangent to the clock face at <math>12</math> o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br />
<br />
<math> \textbf{(A) }\mathrm{2 o'clock} \qquad\textbf{(B) }\mathrm{3 o'clock} \qquad\textbf{(C) }\mathrm{4 o'clock} \qquad\textbf{(D) }\mathrm{6 o'clock} \qquad\textbf{(E) }\mathrm{8 o'clock}</math><br />
<br />
==Solution==<br />
<br />
The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is <math>\boxed{\textbf{(C) }4 \ \mathrm{o'clock}}</math> .<br />
<br />
==Solution 2==<br />
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==Solution 3==<br />
The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is <math>20\pi</math>, so that is <math>20\pi</math> traveled on a <math>60\pi</math> arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_14&diff=67664
2015 AMC 10A Problems/Problem 14
2015-02-06T04:49:34Z
<p>Bob Smith: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externally tangent to the clock face at <math>12</math> o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br />
<br />
[asy]<br />
size(170);<br />
defaultpen(linewidth(0.9)+fontsize(13pt));<br />
draw(unitcircle^^circle((0,1.5),0.5));<br />
path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br />
for(int i=1;i<=12;i=i+1)<br />
{<br />
draw(0.9*dir(90-30*i)--dir(90-30*i));<br />
label("<math>"+(string) i+"</math>",0.78*dir(90-30*i));<br />
}<br />
dot(origin);<br />
draw(shift((0,1.87))*arrow);<br />
draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]<br />
<br />
<math> \textbf{(A) }\mathrm{2 o'clock} \qquad\textbf{(B) }\mathrm{3 o'clock} \qquad\textbf{(C) }\mathrm{4 o'clock} \qquad\textbf{(D) }\mathrm{6 o'clock} \qquad\textbf{(E) }\mathrm{8 o'clock}</math><br />
<br />
==Solution==<br />
<br />
The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is <math>\boxed{\textbf{(C) }4 \ \mathrm{o'clock}}</math> .<br />
<br />
==Solution 2==<br />
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==Solution 3==<br />
The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is <math>20\pi</math>, so that is <math>20\pi</math> traveled on a <math>60\pi</math> arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_14&diff=67663
2015 AMC 10A Problems/Problem 14
2015-02-06T04:49:25Z
<p>Bob Smith: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externally tangent to the clock face at <math>12</math> o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br />
<br />
[asy]<br />
size(170);<br />
defaultpen(linewidth(0.9)+fontsize(13pt));<br />
draw(unitcircle^^circle((0,1.5),0.5));<br />
path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br />
for(int i=1;i<=12;i=i+1)<br />
{<br />
draw(0.9*dir(90-30*i)--dir(90-30*i));<br />
label("<math>"+(string) i+"</math>",0.78*dir(90-30*i));<br />
}<br />
dot(origin);<br />
draw(shift((0,1.87))*arrow);<br />
draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]<br />
<br />
$ \textbf{(A) }\mathrm{2 o'clock} \qquad\textbf{(B) }\mathrm{3 o'clock} \qquad\textbf{(C) }\mathrm{4 o'clock} \qquad\textbf{(D) }\mathrm{6 o'clock} \qquad\textbf{(E) }\mathrm{8 o'clock}<br />
<br />
==Solution==<br />
<br />
The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is <math>\boxed{\textbf{(C) }4 \ \mathrm{o'clock}}</math> .<br />
<br />
==Solution 2==<br />
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==Solution 3==<br />
The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is <math>20\pi</math>, so that is <math>20\pi</math> traveled on a <math>60\pi</math> arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_14&diff=67662
2015 AMC 10A Problems/Problem 14
2015-02-06T04:48:41Z
<p>Bob Smith: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externally tangent to the clock face at <math>12</math> o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br />
<br />
[asy]<br />
size(170);<br />
defaultpen(linewidth(0.9)+fontsize(13pt));<br />
draw(unitcircle^^circle((0,1.5),0.5));<br />
path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br />
for(int i=1;i<=12;i=i+1)<br />
{<br />
draw(0.9*dir(90-30*i)--dir(90-30*i));<br />
label("<math>"+(string) i+"</math>",0.78*dir(90-30*i));<br />
}<br />
dot(origin);<br />
draw(shift((0,1.87))*arrow);<br />
draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]<br />
<br />
==Solution==<br />
<br />
The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is <math>\boxed{\textbf{(C) }4 \ \mathrm{o'clock}}</math> .<br />
<br />
==Solution 2==<br />
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==Solution 3==<br />
The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is <math>20\pi</math>, so that is <math>20\pi</math> traveled on a <math>60\pi</math> arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_14&diff=67661
2015 AMC 10A Problems/Problem 14
2015-02-06T04:48:29Z
<p>Bob Smith: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externally tangent to the clock face at <math>12</math> o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br />
<br />
<math>[asy]<br />
size(170);<br />
defaultpen(linewidth(0.9)+fontsize(13pt));<br />
draw(unitcircle^^circle((0,1.5),0.5));<br />
path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br />
for(int i=1;i<=12;i=i+1)<br />
{<br />
draw(0.9*dir(90-30*i)--dir(90-30*i));<br />
label("</math>"+(string) i+"<math>",0.78*dir(90-30*i));<br />
}<br />
dot(origin);<br />
draw(shift((0,1.87))*arrow);<br />
draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]</math><br />
<br />
==Solution==<br />
<br />
The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is <math>\boxed{\textbf{(C) }4 \ \mathrm{o'clock}}</math> .<br />
<br />
==Solution 2==<br />
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==Solution 3==<br />
The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is <math>20\pi</math>, so that is <math>20\pi</math> traveled on a <math>60\pi</math> arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_14&diff=67660
2015 AMC 10A Problems/Problem 14
2015-02-06T04:48:00Z
<p>Bob Smith: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externally tangent to the clock face at <math>12</math> o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br />
<br />
[asy]<br />
size(170);<br />
defaultpen(linewidth(0.9)+fontsize(13pt));<br />
draw(unitcircle^^circle((0,1.5),0.5));<br />
path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br />
for(int i=1;i<=12;i=i+1)<br />
{<br />
draw(0.9*dir(90-30*i)--dir(90-30*i));<br />
label("<math>"+(string) i+"</math>",0.78*dir(90-30*i));<br />
}<br />
dot(origin);<br />
draw(shift((0,1.87))*arrow);<br />
draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]<br />
<br />
==Solution==<br />
<br />
The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is <math>\boxed{\textbf{(C) }4 \ \mathrm{o'clock}}</math> .<br />
<br />
==Solution 2==<br />
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==Solution 3==<br />
The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is <math>20\pi</math>, so that is <math>20\pi</math> traveled on a <math>60\pi</math> arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_14&diff=67659
2015 AMC 10A Problems/Problem 14
2015-02-06T04:47:23Z
<p>Bob Smith: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externally tangent to the clock face at <math>12</math> o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br />
<br />
[asy]<br />
size(170);<br />
defaultpen(linewidth(0.9)+fontsize(13pt));<br />
draw(unitcircle^^circle((0,1.5),0.5));<br />
path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br />
for(int i=1;i<=12;i=i+1)<br />
{<br />
draw(0.9*dir(90-30*i)--dir(90-30*i));<br />
label("<math>"+(string) i+"</math>",0.78*dir(90-30*i));<br />
}<br />
dot(origin);<br />
draw(shift((0,1.87))*arrow);<br />
draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]<br />
<br />
==Solution==<br />
<br />
The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is <math>\boxed{\textbf{(C) }4 \ \mathrm{o'clock}}</math> .<br />
<br />
==Solution 2==<br />
The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==Solution 3==<br />
The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so therefore, the disk must travel its circumference before the arrow goes up. Its circumference is <math>20\pi</math>, so that is <math>20\pi</math> traveled on a <math>60\pi</math> arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. <math>\boxed{\textbf{(C)}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_1&diff=67421
2015 AMC 10A Problems/Problem 1
2015-02-04T23:30:15Z
<p>Bob Smith: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math><br />
<br />
<math> \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25</math><br />
<br />
==Solution==<br />
<math>(2^0-1+5^2-0)^{-1}\times5</math><br />
<math>=(1-1+25-0)^{-1}\times5</math><br />
<math>=25^{-1}\times5</math><br />
<math>=\frac{1}{25}\times5</math><br />
<math>=\frac{1}{5}\implies{\boxed{\textbf{(C)}\ \frac{1}{5}}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_1&diff=67419
2015 AMC 10A Problems/Problem 1
2015-02-04T23:29:56Z
<p>Bob Smith: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math><br />
<br />
<math> \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25</math><br />
<br />
==Solution==<br />
<math>(2^0-1+5^2-0)^{-1}\times5</math><br />
<math>=(1-1+25-0)^{-1}\times5</math><br />
<math>=25^{-1}\times5</math><br />
<math>=\frac{1}{25}\times5</math><br />
<math>=\frac{1}{5}\implies{\textbf{(C)}\ \frac{1}{5}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_1&diff=67418
2015 AMC 10A Problems/Problem 1
2015-02-04T23:29:15Z
<p>Bob Smith: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math><br />
<br />
<math> \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25</math><br />
<br />
==Solution==<br />
<math>(2^0-1+5^2-0)^{-1}\times5</math><br />
<math>=(1-1+25-0)^{-1}\times5</math><br />
<math>=25^{-1}\times5</math><br />
<math>=\frac{1}{25}\times5</math><br />
<math>=\frac{1}{5}\implies{\boxed{C}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_1&diff=67417
2015 AMC 10A Problems/Problem 1
2015-02-04T23:29:08Z
<p>Bob Smith: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math><br />
<br />
<math> \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25</math><br />
<br />
==Solution==<br />
<br />
==Solution==<br />
<math>(2^0-1+5^2-0)^{-1}\times5</math><br />
<math>=(1-1+25-0)^{-1}\times5</math><br />
<math>=25^{-1}\times5</math><br />
<math>=\frac{1}{25}\times5</math><br />
<math>=\frac{1}{5}\implies{\boxed{C}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_24&diff=67175
2013 AMC 8 Problems/Problem 24
2015-02-02T05:55:59Z
<p>Bob Smith: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Squares <math>ABCD</math>, <math>EFGH</math>, and <math>GHIJ</math> are equal in area. Points <math>C</math> and <math>D</math> are the midpoints of sides <math>IH</math> and <math>HE</math>, respectively. What is the ratio of the area of the shaded pentagon <math>AJICB</math> to the sum of the areas of the three squares?<br />
<br />
<math> \textbf{(A)}\hspace{.05in}\frac{1}{4}\qquad\textbf{(B)}\hspace{.05in}\frac{7}{24}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{3}\qquad\textbf{(D)}\hspace{.05in}\frac{3}{8}\qquad\textbf{(E)}\hspace{.05in}\frac{5}{12}</math><br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J;<br />
<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--J); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
dot("$A$", A, NW);<br />
dot("$B$", B, NE);<br />
dot("$C$", C, NE);<br />
dot("$D$", D, NW);<br />
dot("$E$", E, NW);<br />
dot("$F$", F, SW);<br />
dot("$G$", G, S);<br />
dot("$H$", H, N);<br />
dot("$I$", I, NE);<br />
dot("$J$", J, SE);<br />
</asy><br />
<br />
==Easiest Solution==<br />
<br />
We can obviously see that the pentagon is made of two congruent triangles. We can fit one triangle into the gap in the upper square. Therefore, the answer is just <math>\frac{1}{3}\implies\boxed{C}</math><br />
<br />
<br />
==Solution 1==<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J,X;<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
X= extension(I,J,A,B);<br />
dot(X,red);<br />
draw(I--X--B,red);<br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--J); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
dot("$A$", A, NW);<br />
dot("$B$", B, NE);<br />
dot("$C$", C, NE);<br />
dot("$D$", D, NW);<br />
dot("$E$", E, NW);<br />
dot("$F$", F, SW);<br />
dot("$G$", G, S);<br />
dot("$H$", H, N);<br />
dot("$I$", I, NE);<br />
label("$X$", X,SE);<br />
dot("$J$", J, SE);</asy><br />
<br />
<br />
First let <math>s=2</math> (where <math>s</math> is the side length of the squares) for simplicity. We can extend <math>\overline{IJ}</math> until it hits the extension of <math>\overline{AB}</math>. Call this point <math>X</math>. The area of triangle <math>AXJ</math> then is <math>\dfrac{3 \cdot 4}{2}</math> The area of rectangle <math>BXIC</math> is <math>2 \cdot 1 = 2</math>. Thus, our desired area is <math>6-2 = 4</math>. Now, the ratio of the shaded area to the combined area of the three squares is <math>\frac{4}{3\cdot 2^2} = \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.<br />
<br />
==Solution 2==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J,X;<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
X= (1.25,1);<br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--J); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
dot(X,red);<br />
label("$A$", A, NW);<br />
label("$B$", B, NE);<br />
label("$C$", C, NE);<br />
label("$D$", D, NW);<br />
label("$E$", E, NW);<br />
label("$F$", F, SW);<br />
label("$G$", G, S);<br />
label("$H$", H, N);<br />
label("$I$", I, NE);<br />
label("$X$", X,SW,red);<br />
label("$J$", J, SE);</asy><br />
<br />
Let the side length of each square be <math>1</math>.<br />
<br />
Let the intersection of <math>AJ</math> and <math>EI</math> be <math>X</math>.<br />
<br />
Since <math>[ABCD]=[GHIJ]</math>, <math>AD=IJ</math>. Since <math>\angle IXJ</math> and <math>\angle AXD</math> are vertical angles, they are congruent. We also have <math>\angle JIH\cong\angle ADC</math> by definition. <br />
<br />
So we have <math>\triangle ADX\cong\triangle JIX</math> by <math>\textit{AAS}</math> congruence. Therefore, <math>DX=JX</math>.<br />
<br />
Since <math>C</math> and <math>D</math> are midpoints of sides, <math>DH=CJ=\dfrac{1}{2}</math>. This combined with <math>DX=JX</math> yields <math>HX=CX=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}</math>.<br />
<br />
The area of trapezoid <math>ABCX</math> is <math>\dfrac{1}{2}(AB+CX)(BC)=\dfrac{1}{2}\times \dfrac{5}{4}\times 1=\dfrac{5}{8}</math>.<br />
<br />
The area of triangle <math>JIX</math> is <math>\dfrac{1}{2}\times XJ\times IJ=\dfrac{1}{2}\times \dfrac{3}{4}\times 1=\dfrac{3}{8}</math>.<br />
<br />
So the area of the pentagon <math>AJICB</math> is <math>\dfrac{3}{8}+\dfrac{5}{8}=1</math>.<br />
<br />
The area of the <math>3</math> squares is <math>1\times 3=3</math>.<br />
<br />
Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.<br />
<br />
==Solution 3==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J,K;<br />
A = (0.5,2);<br />
B = (1.5,2);<br />
C = (1.5,1);<br />
D = (0.5,1);<br />
E = (0,1);<br />
F = (0,0);<br />
G = (1,0);<br />
H = (1,1);<br />
I = (2,1);<br />
J = (2,0); <br />
K= (1.25,1);<br />
draw(A--B); <br />
draw(C--B); <br />
draw(D--A); <br />
draw(F--E); <br />
draw(I--J); <br />
draw(J--F); <br />
draw(G--H); <br />
draw(A--J); <br />
filldraw(A--B--C--I--J--cycle,grey);<br />
draw(E--I);<br />
dot(K,red);<br />
label("$A$", A, NW);<br />
label("$B$", B, NE);<br />
label("$C$", C, NE);<br />
label("$D$", D, NW);<br />
label("$E$", E, NW);<br />
label("$F$", F, SW);<br />
label("$G$", G, S);<br />
label("$H$", H, N);<br />
label("$I$", I, NE);<br />
label("$K$", K,SW,red);<br />
label("$J$", J, SE);</asy><br />
<br />
Let the intersection of <math>AJ</math> and <math>EI</math> be <math>K</math>.<br />
<br />
Now we have <math>\triangle ADK</math> and <math>\triangle KIJ</math>.<br />
<br />
Because both triangles has a side on congruent squares therefore <math>AD \cong IJ</math>.<br />
<br />
Because <math>\angle AKD</math> and <math>\angle JKI</math> are vertical angles <math>\angle AKD \cong \angle JKI</math>.<br />
<br />
Also both <math>\angle ADK</math> and <math>\angle JIK</math> are right angles so <math>\angle ADK \cong \angle JIK</math>.<br />
<br />
Therefore by AAS(Angle, Angle, Side) <math>\triangle ADK \cong \triangle KIJ</math>.<br />
<br />
Then translating/rotating the shaded <math>\triangle JIK</math> into the position of <math>\triangle ADK</math><br />
<br />
So the shaded area now completely covers the square <math>ABCD</math><br />
<br />
Set the area of a square as <math>x</math><br />
<br />
Therefore, <math>\frac{x}{3x}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.<br />
<br />
<br />
==See Also==<br />
{{AMC8 box|year=2013|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_25&diff=66749
2008 AMC 12B Problems/Problem 25
2015-01-04T05:06:39Z
<p>Bob Smith: /* Solution */</p>
<hr />
<div>==Problem 25==<br />
Let <math>ABCD</math> be a trapezoid with <math>AB||CD, AB=11, BC=5, CD=19,</math> and <math>DA=7</math>. Bisectors of <math>\angle A</math> and <math>\angle D</math> meet at <math>P</math>, and bisectors of <math>\angle B</math> and <math>\angle C</math> meet at <math>Q</math>. What is the area of hexagon <math>ABQCDP</math>?<br />
<br />
<math>\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \textbf{(E)}\ 36\sqrt{3}</math><br />
<br />
==Solution==<br />
<br />
<center>[[File:2008_AMC_12B_25.jpg]]</center><br />
<br />
Drop perpendiculars to <math>CD</math> from <math>A</math> and <math>B</math>, and call the intersections <math>X,Y</math> respectively. Now, <math>DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2</math> and <math>DX+CY=19-11=8</math>. Thus, <math>DX-CY=3</math>.<br />
We conclude <math>DX=\frac{11}{2}</math> and <math>CY=\frac{5}{2}</math>.<br />
To simplify things even more, notice that <math>90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD</math>, so <math>\angle P=\angle Q=90^{\circ}</math>.<br />
<br />
Also, <cmath>\sin(\angle PDA)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}</cmath><br />
So the area of <math>\triangle APD</math> is: <cmath>R\cdot c\sin a\sin b =\frac{7\cdot7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}</cmath><br />
<br />
Over to the other side: <math>\triangle BCY</math> is <math>30-60-90</math>, and is therefore congruent to <math>\triangle BCQ</math>. So <math>[BCQ]=\frac{5\cdot5\sqrt{3}}{8}</math>.<br />
<br />
The area of the hexagon is clearly <math>[ABCD]-([BCQ]+[APD])</math><cmath>=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}=30\sqrt{3}\implies\boxed{B}</cmath><br />
<br />
==Alternate Solution==<br />
<center>[[File:2008_AMC_12B_25_II.JPG]]</center><br />
<br />
Let <math>AP</math> and <math>BQ</math> meet <math>CD</math> at <math>X</math> and <math>Y</math>, respectively.<br />
<br />
Since <math>\angle APD=90^{\circ}</math>, <math>\angle ADP=\angle XDP</math>, and they share <math>DP</math>, triangles <math>APD</math> and <math>XPD</math> are congruent.<br />
<br />
By the same reasoning, we also have that triangles <math>BQC</math> and <math>YQC</math> are congruent.<br />
<br />
Hence, we have <math>[ABQCDP]=[ABYX]+\frac{[ABCD]-[ABYX]}{2}=\frac{[ABCD]+[ABYX]}{2}</math>.<br />
<br />
If we let the height of the trapezoid be <math>x</math>, we have <math>[ABQCDP]=\frac{\frac{11+19}{2}\cdot x+\frac{11+7}{2}\cdot x}{2}=12x</math>.<br />
<br />
Thusly, if we find the height of the trapezoid and multiply it by 12, we will be done.<br />
<br />
Let the projections of <math>A</math> and <math>B</math> to <math>CD</math> be <math>A'</math> and <math>B'</math>, respectively.<br />
<br />
We have <math>DA'+CB'=19-11=8</math>, <math>DA'=\sqrt{DA^2-AA'^2}=\sqrt{49-x^2}</math>, and <math>CB'=\sqrt{CB^2-BB'^2}=\sqrt{25-x^2}</math>.<br />
<br />
Therefore, <math>\sqrt{49-x^2}+\sqrt{25-x^2}=8</math>. Solving this, we easily get that <math>x=\frac{5\sqrt{3}}{2}</math>.<br />
<br />
Multiplying this by 12, we find that the area of hexagon <math>ABQCDP</math> is <math>30\sqrt{3}</math>, which corresponds to answer choice <math>\boxed{B}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_12&diff=66748
2008 AMC 12B Problems/Problem 12
2015-01-04T05:05:00Z
<p>Bob Smith: /* Alternate Solution */</p>
<hr />
<div>==Problem 12==<br />
For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the <math>2008</math>th term of the sequence?<br />
<br />
<math>\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064</math><br />
<br />
==Solution==<br />
Letting <math>S_n</math> be the nth partial sum of the sequence:<br />
<br />
<math>\frac{S_n}{n} = n</math><br />
<br />
<math>S_n = n^2</math><br />
<br />
The only possible sequence with this result is the sequence of odd integers.<br />
<br />
<math>a_n = 2n - 1</math><br />
<br />
<math>a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}</math><br />
<br />
<br />
==Alternate Solution==<br />
<br />
Letting the sum of the sequence equal <math>a_1+a_2+\cdots+a_n</math> yields the following two equations:<br />
<br />
<math>\frac{a_1+a_2+\cdots+a_{2008}}{2008}=2008</math> and<br />
<br />
<math>\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007</math>.<br />
<br />
Therefore:<br />
<br />
<math>a_1+a_2+\cdots+a_{2008}=2008^2</math> and <math>a_1+a_2+\cdots+a_{2007}=2007^2</math><br />
<br />
Hence, by substitution, <math>a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{B}</math><br />
<br />
(Solution by Bob_Smith)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_12&diff=66747
2008 AMC 12B Problems/Problem 12
2015-01-04T05:04:15Z
<p>Bob Smith: /* Alternate Solution */</p>
<hr />
<div>==Problem 12==<br />
For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the <math>2008</math>th term of the sequence?<br />
<br />
<math>\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064</math><br />
<br />
==Solution==<br />
Letting <math>S_n</math> be the nth partial sum of the sequence:<br />
<br />
<math>\frac{S_n}{n} = n</math><br />
<br />
<math>S_n = n^2</math><br />
<br />
The only possible sequence with this result is the sequence of odd integers.<br />
<br />
<math>a_n = 2n - 1</math><br />
<br />
<math>a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}</math><br />
<br />
<br />
==Alternate Solution==<br />
<br />
Letting the sum of the sequence equal <math>a_1+a_2+\cdots+a_n</math> yields the following two equations:<br />
<br />
<math>\frac{a_1+a_2+\cdots+a_{2008}}{2008}=2008</math> and<br />
<br />
<math>\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007</math>.<br />
<br />
So then:<br />
<br />
<math>a_1+a_2+\cdots+a_{2008}=2008^2</math> and <math>a_1+a_2+\cdots+a_{2007}=2007^2</math><br />
<br />
Therefore, by substitution, <math>a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{B}</math><br />
<br />
(Solution by Bob_Smith)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_12&diff=66746
2008 AMC 12B Problems/Problem 12
2015-01-04T05:03:22Z
<p>Bob Smith: /* Alternate Solution */</p>
<hr />
<div>==Problem 12==<br />
For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the <math>2008</math>th term of the sequence?<br />
<br />
<math>\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064</math><br />
<br />
==Solution==<br />
Letting <math>S_n</math> be the nth partial sum of the sequence:<br />
<br />
<math>\frac{S_n}{n} = n</math><br />
<br />
<math>S_n = n^2</math><br />
<br />
The only possible sequence with this result is the sequence of odd integers.<br />
<br />
<math>a_n = 2n - 1</math><br />
<br />
<math>a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}</math><br />
<br />
<br />
==Alternate Solution==<br />
<br />
Letting the sum of the sequence equal <math>a_1+a_2+\cdots+a_n</math> yields the following two equations:<br />
<br />
<math>\frac{a_1+a_2+\cdots+a_{2008}}{2008}=2008</math> and<br />
<br />
<math>\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007</math>.<br />
<br />
So then:<br />
<br />
<math>a_1+a_2+\cdots+a_{2008}=2008^2</math> and <math>a_1+a_2+\cdots+a_{2007}=2007^2</math><br />
<br />
Therefore, by substitution, <math>a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{B}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_12&diff=66745
2008 AMC 12B Problems/Problem 12
2015-01-04T05:01:08Z
<p>Bob Smith: /* Solution */</p>
<hr />
<div>==Problem 12==<br />
For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the <math>2008</math>th term of the sequence?<br />
<br />
<math>\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064</math><br />
<br />
==Solution==<br />
Letting <math>S_n</math> be the nth partial sum of the sequence:<br />
<br />
<math>\frac{S_n}{n} = n</math><br />
<br />
<math>S_n = n^2</math><br />
<br />
The only possible sequence with this result is the sequence of odd integers.<br />
<br />
<math>a_n = 2n - 1</math><br />
<br />
<math>a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}</math><br />
<br />
<br />
==Alternate Solution==<br />
<br />
<math>\frac{a_1+a_2+...+a_{2008}}{2008}=2008</math> and<br />
<br />
<math>\frac{a_1+a_2+...+a_{2007}}{2007}=2007</math>.<br />
<br />
So then:<br />
<br />
<math>a_1+a_2+...+a_{2008}=2008^2</math> and <math>a_1+a_2+...+a_{2007}=2007^2</math><br />
<br />
Therefore, by substitution, <math>a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{B}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>
Bob Smith
https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_25&diff=66744
2008 AMC 12B Problems/Problem 25
2015-01-04T04:52:20Z
<p>Bob Smith: /* Solution */</p>
<hr />
<div>==Problem 25==<br />
Let <math>ABCD</math> be a trapezoid with <math>AB||CD, AB=11, BC=5, CD=19,</math> and <math>DA=7</math>. Bisectors of <math>\angle A</math> and <math>\angle D</math> meet at <math>P</math>, and bisectors of <math>\angle B</math> and <math>\angle C</math> meet at <math>Q</math>. What is the area of hexagon <math>ABQCDP</math>?<br />
<br />
<math>\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \textbf{(E)}\ 36\sqrt{3}</math><br />
<br />
==Solution==<br />
<br />
<center>[[File:2008_AMC_12B_25.jpg]]</center><br />
<br />
Drop perpendiculars to <math>CD</math> from <math>A</math> and <math>B</math>, and call the intersections <math>X,Y</math> respectively. Now, <math>DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2</math> and <math>DX+CY=19-11=8</math>. Thus, <math>DX-CY=3</math>.<br />
We conclude <math>DX=\frac{11}{2}</math> and <math>CY=\frac{5}{2}</math>.<br />
To simplify things even more, notice that <math>90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD</math>, so <math>\angle P=\angle Q=90^{\circ}</math>.<br />
<br />
Also, <cmath>\sin(\angle PDA)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}</cmath><br />
So the area of <math>\triangle APD</math> is: <cmath>R\cdot c\sin a\sin b =\frac{7\cdot7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}</cmath><br />
<br />
Over to the other side: <math>\triangle BCY</math> is <math>30-60-90</math>, and is therefore congruent to <math>\triangle BCQ</math>. So <math>[BCQ]=\frac{5\cdot5\sqrt{3}}{8}</math>.<br />
<br />
The area of the hexagon is clearly <math>[ABCD]-([BCQ]+[APD])</math><cmath>=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}=30\sqrt{3},\implies\boxed{B}</cmath><br />
<br />
==Alternate Solution==<br />
<center>[[File:2008_AMC_12B_25_II.JPG]]</center><br />
<br />
Let <math>AP</math> and <math>BQ</math> meet <math>CD</math> at <math>X</math> and <math>Y</math>, respectively.<br />
<br />
Since <math>\angle APD=90^{\circ}</math>, <math>\angle ADP=\angle XDP</math>, and they share <math>DP</math>, triangles <math>APD</math> and <math>XPD</math> are congruent.<br />
<br />
By the same reasoning, we also have that triangles <math>BQC</math> and <math>YQC</math> are congruent.<br />
<br />
Hence, we have <math>[ABQCDP]=[ABYX]+\frac{[ABCD]-[ABYX]}{2}=\frac{[ABCD]+[ABYX]}{2}</math>.<br />
<br />
If we let the height of the trapezoid be <math>x</math>, we have <math>[ABQCDP]=\frac{\frac{11+19}{2}\cdot x+\frac{11+7}{2}\cdot x}{2}=12x</math>.<br />
<br />
Thusly, if we find the height of the trapezoid and multiply it by 12, we will be done.<br />
<br />
Let the projections of <math>A</math> and <math>B</math> to <math>CD</math> be <math>A'</math> and <math>B'</math>, respectively.<br />
<br />
We have <math>DA'+CB'=19-11=8</math>, <math>DA'=\sqrt{DA^2-AA'^2}=\sqrt{49-x^2}</math>, and <math>CB'=\sqrt{CB^2-BB'^2}=\sqrt{25-x^2}</math>.<br />
<br />
Therefore, <math>\sqrt{49-x^2}+\sqrt{25-x^2}=8</math>. Solving this, we easily get that <math>x=\frac{5\sqrt{3}}{2}</math>.<br />
<br />
Multiplying this by 12, we find that the area of hexagon <math>ABQCDP</math> is <math>30\sqrt{3}</math>, which corresponds to answer choice <math>\boxed{B}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>
Bob Smith