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!
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!
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.
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.
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.
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