PointXbelongs to the interior of a triangleABC. The distance from this point to linesAB, BC, ACisk, l, mrespectively andris the length of an inradius of this triangle. Prove that:

$\displaystyle 2r<k+l+m$

Thanks in advance!

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- Jan 29th 2011, 09:44 AMterrogeometric proof - triangle
Point

*X*belongs to the interior of a triangle*ABC*. The distance from this point to lines*AB, BC, AC*is*k, l, m*respectively and*r*is the length of an inradius of this triangle. Prove that:

$\displaystyle 2r<k+l+m$

Thanks in advance! - Feb 2nd 2011, 08:38 PMabhishekkgp
well my proof is not completely geometric but still it may help.

its easy to see that $\displaystyle \frac{1}{2}ck + \frac{1}{2}la +\frac{1}{2}mb = \Delta$

$\displaystyle \rightarrow ck+la+mb=2\Delta$

$\displaystyle \rightarrow \frac{ck+la+mb}{s}=\frac{2\Delta}{s}=2r$

now it only remains to prove that $\displaystyle (k+l+m)s>ck+la+mb$

thia is very easy. give it a try. - Feb 7th 2011, 05:03 PMJoshuaJava
Okay maybe I'm crazy but does this prove it? :

If point X is at the centre of the incircle then the distances from this point to each line are the shortest possible and are also equal to r. Therefore, k + l + m = 3r ..... therefore k + l + m > 2r

Somebody tell me if this is right please, and if it is not, please tell me why. Thanks. - Feb 7th 2011, 07:08 PMabhishekkgp
- Feb 8th 2011, 01:09 AMFlatiron
- Feb 8th 2011, 08:35 AMJoshuaJava
Because k l and m are at the centre of the incircle.

- Feb 8th 2011, 08:38 AMJoshuaJava
It does cover the general case because if the point is at the centre of the incircle, it means that if it were anywhere else, the sum of each length would be greater than the sum of each length from the cente, therefore it would definately be greater than 2r at any other point in the triangle.

- Feb 8th 2011, 08:41 AMabhishekkgp
- Feb 8th 2011, 09:54 AMArchie Meade
At the incentre, the triangle area can be expressed as the sum of the areas of 3 internal triangles

$\displaystyle A=\displaystyle\frac{1}{2}r(a+b+c)$

At the point, the triangle area can be expressed as the sum of the areas of 3 other internal triangles

$\displaystyle A=\displaystyle\frac{1}{2}kb+\frac{1}{2}la+\frac{1 }{2}mc=\frac{1}{2}(kb+la+mc)$

Then

$\displaystyle r(a+b+c)=kb+la+mc$

$\displaystyle 2r=\displaystyle\frac{2(kb+la+mc)}{a+b+c}$

$\displaystyle \displaystyle\frac{2(kb+la+mc)}{a+b+c}<k+l+m\;\;?$

$\displaystyle 2(kb+la+mc)<(k+l+m)(a+b+c)\;\;?$

$\displaystyle 2kb+2la+2mc<(a+b+c)k+(a+b+c)l+(a+b+c)m\;\;?$

Since the length of any side of a triangle is less than the sum of the other two sides, then

$\displaystyle a+b+c>2b,\;\;a+b+c>2a,\;\;a+b+c>2c$

Therefore the inequality is true - Feb 9th 2011, 07:34 AMJoshuaJava
Damn... I don't know. But I'm sure there is a way to prove it. Sorry for wasting your time though.

- Feb 9th 2011, 10:54 AMArchie Meade
- Feb 13th 2011, 08:36 AMJoshuaJava
Yes, but, as I said, if it is at the incentre, the sum of the lengths is smaller than anywhere else in the triangle, but I don't know how to prove this, but it is true.

- Feb 13th 2011, 08:54 AMArchie Meade
The incentre is one of infinitely many points in the triangle.

There is no reference to it in the original post at the start of the thread.