me too. You are dividing by c, which is less than 13b/6. The result will be larger than 42/13. What do you think of my attempt on the question as posted earlier?
thanks a lot guys for your interest. Yes, it is somewhat of a confusing problem. @Steve: I appreciated it a lot and you did a very good job at explaining it.
Thanks a lot guys and Happy New year. take care!
In a right triangle, if the legs (altitudes are 12 and 14), then
if we pick an angle θ and let h be the altitude to the hypotenuse,
h/14 = sinθ
h/12 = cosθ
so h^2/144 + h^2/196 = 1, and h =~ 9.1
So, Reiny's estimate is way low. Now the question is, does a right triangle provide maximum 3rd altitude h?
More thought needed.
Take a look at
where this very question is discussed.
Just a final note. Since the 3rd altitude h is bounded by
1/a + 1/b <= 1/h <= 1/a - 1/b
where a < b,
the limiting case where a=b (an isosceles triangle) shows that
1/a+1/a <= 1/h <= 1/a - 1/a
2/a <= 1/h <= 0
shows that h >= a/2, and may grow without bound.
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