## Thursday, January 19, 2017

### Geometry Problem 1307 Triangle, Incenter, Parallel line, Sides, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Let h be the length of the altitude from B.
From similar triangles,
r/MI = h/c and area of Tr. ABC, S = r.s where r is the inradius and s the semi-perimeter and moreover,
S = 1/2 hb
These equations yield MI = 1/2bc/s = bc/(a+b+c)
Similarly NI = ba/(a+b+c)

So MN = MI+NI = b(a+c)/(a+b+c)

Sumith Peiris
Moratuwa
Sri Lanka

2. Problem 1307 - Alternate Solution

Let h be the height of the altitude from B and r the inradius and s the semi perimeter,
Let S(ABC) = ∆.

S(BMN)/∆= MN2/b2 = ½ MN(h-r)/(½bh)

Since h=2∆/s so

MN = b(h-r)/h = b(2∆/b - r)/(2∆/b)
= b(2∆ - br)/2∆
= b(2rs – br)/2rs
= b(2s– b)/2s
= b(a+c)/(a+b+c)

Sumith Peiris
Moratuwa
Sri Lanka

3. Problem 1037
Let (ABC) is area the triangle ABC then (ABC)=sr=bh/2.( 2s=a+b+c,r=radius of the inscribed circle and h=altitude of the triangle ABC).So h=2sr/b=(a+b+c)r/b. But MN//AC
then from Thales' Theorem and triangle BMN with triangle ABC are similar we have
MN/AC=h1/h=(h-r)/h=1-r/h=1-b/(a+b+c)=(a+c)/(a+b+c).Therefore MN=b(a+c)/(a+b+c).
APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

4. Let h and r are height from B and inradius of triangle ABC
We have MN/AC= (h-r)/h= 1- r/h
2 * area of ABC= b.h=2p.r
So r/h= b/2p= b/(a+b+c)
So MN=b*(1-r/h)= b*(a+c)/(a+b+c)

5. Draw line from the point M perpendicular onto AC. The line meets AC at the point P
Draw line from the point N perpendicular onto AC. The line meets AC at the point Q

Let K be the point of tangency of the circle I with AB.
Let H be the point of tangency of the circle I with BC.

Then we see that the triangle AMP is congruent with MIK , and the the triangle CNQ is congruent with NIH.

Therefore AM+CN =MN.

Triangle BMN is similar to BAC therefore :

c-AM/MN=c/b
(1.)AMb=bc-cMN

and

a-CN/MN=a/b
ab-CNb=MNa
(2.)CNb=ab-MNa

Adding equations 1 and 2 with CN+AM=MN

b(CN+AM)=ab-MNa+bc-cMN
bMN=ab+bc-MN(a+c)
MN(a+b+c)=ab+bc
MN=b(a+c)/(a+b+c)