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.
+129849 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
+1490 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 
