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January 24, 2015

January 24, 2015

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

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 12:21pmme 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 1:36pmthanks 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 4:28pmIn 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 5:09pmTake 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 4:54amJust 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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