All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Median Maximum (Posted on 2008-02-03)
Two medians of a triangle have lengths of 12 and 15. How long is the third median when the area of the triangle is a maximum? (Try to solve this without calculus.)

 See The Solution Submitted by Brian Smith Rating: 3.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 1 of 3
`Let ABC be the triangle, X' the midpoint of the sideopposite vertex X, and AA' and BB' the given medians.`
`If the coordinates of the centroid are (0,0), thenthe coordinates of the vertices and midpoints are`
`   A = (2a,0)                      A' = (-a,0)`
`   B = (2b*cos(x),2b*sin(x))       B' = (-b*cos(x),-b*sin(x))`
`   C = (-2b*cos(x)-2a,-2b*sin(x))  C' = (b*cos(x)+a,b*sin(x))`
`Therefore,`
`   Area(ABC) = Area(ABA') + Area(ACA')`
`             = 2*Area(ABA')`
`             = 6ab*sin(x)`
`Clearly, the maximum area occurs when x = 90 degrees.`
`Therefore,`
`   C = (-2a,-2b)   C' = (a,b)`
`               and`
`   |CC'| = 3*sqrt(a^2+b^2)`
`For our problem,`
`   |CC'| = 3*sqrt(5^2+4^2) = 3*sqrt(41)`
` `

 Posted by Bractals on 2008-02-03 12:13:42

 Search: Search body:
Forums (0)