# geometric proof - triangle

• Jan 29th 2011, 10:44 AM
terro
geometric 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:

$2r

• Feb 2nd 2011, 09:38 PM
abhishekkgp
Quote:

Originally Posted by terro
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:

$2r

well my proof is not completely geometric but still it may help.

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

now it only remains to prove that $(k+l+m)s>ck+la+mb$
thia is very easy. give it a try.
• Feb 7th 2011, 06:03 PM
JoshuaJava
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, 08:08 PM
abhishekkgp
Quote:

Originally Posted by JoshuaJava
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.

how do you have k+l+m=3r???
• Feb 8th 2011, 02:09 AM
Flatiron
Quote:

Originally Posted by JoshuaJava
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 is correct for the special case where point X is the center of the circle, but the proof should cover the general case where X is any point in the interior of the triangle.
• Feb 8th 2011, 09:35 AM
JoshuaJava
Because k l and m are at the centre of the incircle.
• Feb 8th 2011, 09:38 AM
JoshuaJava
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, 09:41 AM
abhishekkgp
Quote:

Originally Posted by JoshuaJava
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 centre.

how do you prove this??
• Feb 8th 2011, 10:54 AM
Quote:

Originally Posted by terro
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:

$2r

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

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

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

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

Then

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

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

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

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

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

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

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

Therefore the inequality is true
• Feb 9th 2011, 08:34 AM
JoshuaJava
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, 11:54 AM
Quote:

Originally Posted by JoshuaJava
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 doesn't work because the point can be anywhere inside the triangle.
Therefore one of the lengths could be very close to zero.
• Feb 13th 2011, 09:36 AM
JoshuaJava
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, 09:54 AM