Posted by **Reiny** on Tuesday, January 1, 2013 at 11:49am.

have been looking at your post ...

http://www.jiskha.com/display.cgi?id=1357047578
interesting question...

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

http://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.577996.html

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.

## Answer this Question

## Related Questions

- Spanish for Alicia and/or Vicky - Be sure you see all 3 of your posts: 1. http...
- To Kuromi - re English - http://www.jiskha.com/display.cgi?id=1254257821 and ...
- Arts/social science - What is the relevance of academic study of the Bible today...
- written communication - What steps would one take to ensure that the purpose of ...
- Written Communication - Discuss how you would approach creating a negative ...
- English for sou - I hope you're still working on these: http://www.jiskha.com/...
- To Al 92 - http://www.jiskha.com/display.cgi?id=1278284743 I haven't been on in ...
- Two for Damon - I don't know the rowing terminology! http://www.jiskha.com/...
- Spanish III for Whitney - I have a suggestion for you on the last post of Dec. 9...
- Spanish/History for Laura & Carry - Thank you for using the Jiskha Homework Help...

More Related Questions