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In △ABC, AB=6, BC=4 and AC=8. A segment parallel to BC and tangent to the incircle of △ABC intersects AB at M and AC at N.  Find MN.

May 22, 2020

#1
+111456
+2

We  can find  the height  of  this triangle  as  follows

Using  Heron's Formula  we have  the area as

sqrt  (9 (9-8) (9-6) (9 - 4))  = sqrt  ( 9 * 3 * 5)  =  3sqrt(15)

So.....to find the height we  have

Area  = (1/2)(BC) height

3sqrt (15)  =  (1/2)(4) height

3sqrt (15)  = 2 * height

(3/2)sqrt (15)  = 1.5 sqrt (15)  =  height  = y coordinate of A

And we  can find  the  x coordinate  of A  by the Pythagorean Theorem

sqrt  [ AC^2  -  height of ABC^2 ]   = sqrt  (8^2   -(1.5 sqrt (15))^2  )  = sqrt (30.25)  =  5.5

So....the coordinates  of A  = (5.5, 1.5sqrt (15))

We  can use a formula to find  the coordinates  of  the center  of the incircle

Let B = (4,0)  and C  = (4,0)

x coordinate  of incenter  =

[ Ax* a  + Bx* b  + Cx * c ]  / perimeter

Where Ax = the x coordinate  of A  Bx  = x coordinate of B    Cx  = x coordinate of C

And a, b , c  are  the sides lengths opposite A, B and C

So  we have

[ 5.5 (4)  + (4)(8)  + (0)(6)] / 18  =  3

Similarly the  y coordinate  of the center of  the incircle  =

[Ay * a  + By * b + Cy * c ]  / 18

[ 1.5sqrt (15)(4)  + (0)(8) + (0)(6)  / 18   =  sqrt (15)/3   =    radius  of incircle

Then the  height of triangle ANM = height of triangle ABC  - 2* incircle radius = sqrt (15)  ( 3/2 - 2/3)  =

(5/6) sqrt (15 )

And since triangles  AMN and ABC  are similar

Then

MN  / height of AMN  =  BC / height of ABC

MN  / [ (5/6)sqrt (15) ] =   4 /  [  (3/2 ) sqrt (15) ]

MN =  4 (5/6)  / (3/2)   =  4 ( 5/6) (2/3)  =  40 /18  =  20/9

May 22, 2020
#2
+1326
+2

Using the law of cosines we can find angles B and C.

∠B = 104.478°             ∠OBL = ∠B /2 = 52.239°

∠C = 46.567°               ∠OCL = ∠C /2 = 23.2835°

Using the law of sines we can find the side CO of the triangle COB.

CO = 3.266

Radius   r = sin(OCL) * CO = 1.291

NQ = MR = 2r = 2.582

CQ = NQ / tan(∠C) = 2.444

BQ = 4 - CQ = 1.556

Angle MBR = 180 - ∠B = 75.522°

BR = MR / tan(MBR) = 0.667

And finally...       MN = BQ + BR = 2.223

May 22, 2020