Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Thursday, January 19, 2017

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

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geometry problem,
incenter,
measurement,
parallel,
side,
triangle

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Let h be the length of the altitude from B.

ReplyDeleteFrom 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

Problem 1307 - Alternate Solution

ReplyDeleteLet 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

Problem 1037

ReplyDeleteLet (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

Let h and r are height from B and inradius of triangle ABC

ReplyDeleteWe 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)

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

ReplyDeleteDraw 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)