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for Knights

posted by .

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 -

    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 -

    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 -

    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 -

    Take a look at

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

    where this very question is discussed.

  • for Knights -

    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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