Question

Using Triangle Inequalities

Answer 1

The lengths of two altitudes of a triangle are h and k,where \(h\ne k\). Determine upper and lower bounds for the length of the third altitude in terms of h and k .

Answer 2

@iambatman @ikram002p @Kainui

Answer 3

The very first thing is:

\[\large{ah = bk = cl}\tag{1}\]

Answer 4

\[\large{\implies \cfrac{a}{c} = \cfrac{l}{h}; \cfrac{b}{c} = \cfrac{l}{k}}\]

Answer 5

WLOG, let h<k

Now using triangle inequality :

\[\large{c<a+b}\]

We obtain:

\[\large{\cfrac{a}{c} <\cfrac{b}{c} + 1}\]

Answer 6

How do we know ah=bk=cl? Originally I had thought you said al=ch+bk since those are all representations of twice the area. But I guess I just don't know triangles really well at all.

Answer 7

Also this implies that:

\[\large{\cfrac{l}{h} < \cfrac{l}{k} + 1}\]

\[\large{\implies l(\cfrac{1}{h} - \cfrac{1}{k}) < 1}\]

\[\large{\implies l < \cfrac{hk}{k-h}}\tag{2}\]

Now again using triangle inequality:

\[\large{c < a+b}\]
We obtain:

\[\large{1 <\cfrac{a}{b} + \cfrac{b}{c}; 1< \cfrac{l}{h} + \cfrac{l}{k}}\]

This implies that:

\[\large{l > \cfrac{hk}{h+k}}\]

Answer 8

@Kainui its a relationship between altitudes of a triangle and the corresponding sides

Answer 9

Thus:

\[\large{\boxed{\cfrac{hk}{k-h}>l>\cfrac{hk}{h+k}}}\tag{ANS}\]

Answer 10

This is how we can use triangle inequalities in inequalities problems. These sort of problems are classified as Geometric Inequalities problems

Answer 11

@Kainui If you are interested I can prove that theorem in 2-3 steps :)

Answer 12

Yes please do. I think everything else makes sense to me though, as I am really comfortable with the triangle inequality haha.

Answer 13

Using sine rule we can obtain the required equation :)