Prove that there is no triangle whose altitudes are of length 4, 7, and 10 units.
Assume the contrary.
Call the sides A,B,C. The area of a triangle = 1/2bh, so area (1/2)*4*A=area(1/2)*7*B=area(1/2)*10*C
so C is the shortest side and A is the longest.
Now let B=(ab), C=(abc), so that (1/2)*4*a+(1/2)*7*(ab)=10*(abc)
Simplifying:
13b+20c=9a, a=(13b+20c)/9
Substituting:
(1/2)*4*((13b+10c)/9) +(1/2)*7*(((13b+10c)/9)b)=10*(((13b+10c)/9)bc)
Simplifying again:
40b+55c=40b+10c, so c = 0.
But this implies that B=C, a contradiction, so there is no such triangle.
Edited on October 4, 2013, 4:35 am

Posted by broll
on 20131004 04:33:00 