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=(a-b), C=(a-b-c), so that (1/2)*4*a+(1/2)*7*(a-b)=10*(a-b-c)

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)-b-c)

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 2013-10-04 04:33:00