April 1st, 2019, 03:55 PM  #11 
Member Joined: Mar 2019 From: TTF area Posts: 83 Thanks: 1 
H is Hamilton, but that's quite irrelevant. I think the plane crashed at (71,25) rounded; is this correct?
Last edited by skipjack; April 1st, 2019 at 04:24 PM. 
April 1st, 2019, 04:10 PM  #12  
Math Team Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545  Quote:
Quote:
What about the altitude from Miami to ... where? Last edited by skipjack; April 1st, 2019 at 04:27 PM.  
April 1st, 2019, 04:18 PM  #13 
Member Joined: Mar 2019 From: TTF area Posts: 83 Thanks: 1 
The altitude part confuses me also. I believe it says on altitude to Miami, so I don't think we have to solve altitude. Perpendicular bisectors are correct; that's how I solved. I found the perpendicular bisector of HS and MH.
Last edited by skipjack; April 1st, 2019 at 04:29 PM. 
April 1st, 2019, 04:24 PM  #14 
Math Team Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545  I thought you said point H was irrelevant?
Last edited by skipjack; April 1st, 2019 at 04:29 PM. 
April 1st, 2019, 04:27 PM  #15 
Member Joined: Mar 2019 From: TTF area Posts: 83 Thanks: 1 
I see how you could get that; I meant what H stands for is irrelevant. Sorry for this confusion. Basically, I found the midpoints of HS and MH, found their slopes/gradients, then did the equations of the gradients. After solving for x, I put x into one of my equations to solve for y. Last edited by skipjack; April 1st, 2019 at 04:57 PM. 
April 1st, 2019, 04:55 PM  #16 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  
April 1st, 2019, 05:06 PM  #17 
Math Team Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545 
You have three noncollinear points, M(80,26) H(65,32) and S(66,18 ), that form a triangle. The perpendicular bisector of segment MS passes through its midpoint, (73,22). The slope of MS is 4/7, therefore, the slope of the perpendicular bisector slope 7/4. The equation of the perpendicular bisector of MS is $y = \dfrac{7}{4}(x + 66) + 18$ The altitude from M to HS passes through M(80,26) and is perpendicular to HS ... The slope of HS is 14, so the altitude to HS has slope 1/14. The equation of the altitude from M to HS is $y = \dfrac{1}{14}(x + 80) + 26$ The intersection of the altitude from M to HS and the perpendicular bisector of MS can be found by setting the two lines equal. You calculated (71,25) rounded. Your yvalue is good ... your xcoordinate is off according to my calculation. 
April 1st, 2019, 05:21 PM  #18 
Math Team Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545  Source for that value?
Last edited by skipjack; April 1st, 2019 at 05:31 PM. 
April 1st, 2019, 05:54 PM  #19  
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  Quote:
I had used an outdated value of 6080 feet for the nautical mile. The currently used value is exactly 1.852 km, so you were correct. According to wikipedia, this value was first defined in 1929 (at an international conference held in Monaco). However, the United States adopted this value in 1954 and Britain adopted it in 1970.  
April 1st, 2019, 06:25 PM  #20  
Math Team Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545  Quote:
perpendicular bisector of MS should be $y = \dfrac{7}{4}(x+73)+22$ making the intersection coordinates approx (71,25) so the OP is correct  

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