Posted by Reiny on Tuesday, January 1, 2013 at 11:49am.
have been looking at your post ...
let the altitude of 12 hit base a
let the altitude of 14 hit base b
then (1/2)(12)a = (1/2)(14)b
6a = 7b
a = 7b/6
let the third side be c and its altitude be x
from the properties of the sides of a triangle
c < a+b
c < 7b/6 + b
c < 13b/6
but by area property
cx = 7b or cx = 6a
take cx = 7b
the absolute maximum value that c could be is a number just a "smidgeon" less than 13b/6 , which is positive , so divide both sides of
cx = 7b by that
cx/c < 7b / (13b/6)
c < 42/13
since you want the largest integer, that would be 3
( I am a little uneasy about my last division of the inequality)
- for Knights - Steve, Tuesday, January 1, 2013 at 12:21pm
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?
- for Knights - Knights, Tuesday, January 1, 2013 at 1:36pm
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!
- for Knights - Steve, Tuesday, January 1, 2013 at 4:28pm
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.
- for Knights - Ta Daaaa - Steve, Tuesday, January 1, 2013 at 5:09pm
Take a look at
where this very question is discussed.
- for Knights - Steve, Wednesday, January 2, 2013 at 4:54am
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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