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 07-02-2005, 12:10 AM #1 Member   Join Date: Oct 2004 Location: Tampa Bay Area Posts: 723 Math CHALLENGE...Post the hardest math SAT I questions you can find.. i could use the practice ....let's see who can solve it the most efficiently...we can all learn from these efficient methods... Reply
 07-02-2005, 12:21 AM #2 Member   Join Date: Jun 2005 Posts: 667 Given a triangle ABC as described below: Side AB = Side AC. Draw a line from C to side AB. call that line CD. Now draw a line from B to side AC. Call that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. The question--find what angle EDC is. (This can be, and is meant to be, solved geometrically, without trig. ) Reply
 07-02-2005, 12:38 AM #3 Junior Member   Join Date: Jun 2005 Posts: 39 is it 10 degrees? LOL just a guess Reply
 07-02-2005, 12:40 AM #4 Member   Join Date: Jun 2005 Posts: 667 No, it is not. Reply
 07-02-2005, 01:00 AM #5 Junior Member   Join Date: Jan 2005 Posts: 169 Is this 75? Reply
 07-02-2005, 01:03 AM #6 Junior Member   Join Date: Nov 2004 Location: San Diego, CA Posts: 207 Is it 60 degrees? Reply
 07-02-2005, 01:11 AM #7 Junior Member   Join Date: Jun 2005 Posts: 39 oops i put my tick mark on the wrong side.. giving me the wrong isoceles!! did it again but i still thinkg i got it wrong .. 65? .. if the mid point (let's say O) is where everythign intersects and BOC = 50 thus DOE = 50 .. (my logic might be wrong but i believe triangle DOE is an isoceles) leave 130 degrees divided by 2.. 65? there must be an easier way ahha EDIT: Vehement answers works out for me.. did that the first time with my wrong isoceles.. but changed my method second time around =/ whoops Last edited by iducky; 07-02-2005 at 01:20 AM. Reply
 07-02-2005, 01:11 AM #8 Member   Join Date: Mar 2005 Posts: 943 double post Reply
 07-02-2005, 01:13 AM #9 Member   Join Date: Mar 2005 Posts: 943 its 70..... ok the full steps first add up bde, ebc, ecb and ecd. you will get 160. 180 - 160 = angel a 180 - angel a(20)/2 = ade = 80 180 - dbe - dbc - bce = angel bdc = 30 subtract bdc(30) and ade(80) from 180 = the answer = 70 Last edited by Vehement; 07-02-2005 at 01:24 AM. Reply
 07-02-2005, 01:24 AM #10 Senior Member   Join Date: Aug 2004 Location: Xiggilandia where the ale trumps Westvleteren Posts: 15,628 ........ Reply
 07-02-2005, 01:37 AM #11 Member   Join Date: Jun 2005 Posts: 667 It is not 70. Reply
 07-02-2005, 01:38 AM #12 Super Moderator   Join Date: Jun 2005 Location: Northeast Posts: 2,013 courtesy requested - please, title your submissions! It'll make it so much easier to jump right to the next question if you are not interested in one before. thanks much! Reply
 07-02-2005, 01:41 AM #13 Member   Join Date: Mar 2005 Posts: 943 Hmm you copied the question wrong then? or i seriously misread. i have figured out every angle in the triangle so i really don't know where my prob is Reply
 07-02-2005, 01:41 AM #14 Junior Member   Join Date: Jun 2005 Posts: 153 I messed up Reply
 07-02-2005, 01:49 AM #15 Senior Member   Join Date: Aug 2004 Location: Xiggilandia where the ale trumps Westvleteren Posts: 15,628 Vehement, the problem is far from being a sixty seconds SAT question. It is the Gruber's challenge. It takes a ... long time to develop a solution. Reply

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