Question

hi guys ive be struggling on this problem for a couple of days, so please help if you can


Problem:Two altitudes of a triangle have lengths 12 and 14. What is the longest possible length of the third altitude, if it is a positive integer?

Thanks in advance and merry christmas: happy new year!

Answer 1

I'm not sure there is one. Consider an isosceles triangle, where the two shorter altitudes are the same, say h.

If the distance from the base of the altitude to a base vertex is c, then the base angle θ is such that

tanθ = h/c
If the remaining altitude is k, then we also have

tanθ = k/(√(h^2+c^2)/2) = h/c
k = h√(h^2+c^2)/2c

as c gets smaller and smaller, k gets closer and closer to h^2/2c, which can grow without limit.

Granted, we don't have an isosceles triangle, but the two altitudes are almost equal, and I think the same argument could be made.

Answer 2

I tried a different approach, but I don't like my last sequence of steps.
What is your opinion ?
http://www.jiskha.com/display.cgi?id=1357058982

I even tried playing with the triangle here ...

http://www.mathopenref.com/triangleorthocenter.html

Answer 3

My argument above fails because we don't have an isosceles triangle. In the limit, with an isosceles triangle, we end up with an infinitely tall "triangle" with parallel sides h units apart. However, because we have diagonals 12 and 14, that is not possible.

More thought required.

Answer 4

Two altitudes of a triangle have lengths 12 and 14. What is the longest possible length of the third altitude, if it is a positive integer?