Problem:
Point D lies on side BC of β³ABC so that AD bisects β BAC. The perpendicular bisector of AD intersects the bisectors of β ABC and β ACB in points E and F, respectively. Given that AB=4,BC=5, and CA=6, the area of β³AEF can be written as pmnββ, where m and p are relatively prime positive integers, and n is a positive integer not divisible by the square of any prime. Find m+n+p.
Solution:
Denote MAβ,MBβ,MCβ to be the midpoints of the arc BC, CA, AB.
Notice by the incenter excenter lemma, that IMAβ=MAβB=MAβC.
Then, using Ptolemy's on ABCMAβ yields
BCβ AMAβ=ABβ CMAβ+ACβ MAβB
This yields that 2MAβI=MAβAβMAβI=IA
Notice that
DMAβDIβ=BMAβsinβ CBMAβBIsinβ IBCβ
which can be simplified to
DMAβDIβ=BMAβsin2β AβBIsin2β Bββ=1
And so D is a midpoint of IMAβ.
Since AI gets perpendicularly bisected by MBβMCβ, and AD is perpendicularly bisected by EF, and that D is a midpoint of AI, we see that there's a homothety of DEF and ABC with ratio 21β around I.
This means that [AEF]=[DEF]=(21β)2MAβMBβMCβ.
We can find the circumradius that R=787ββ.
Notice the area of a triangle can be written as
[ABC]=4Rabcβ
Rewriting
R=sinAaβ=sinBbβ=sinCcβ
or
a=RsinA,b=RsinB,c=RsinC
This yields that
[ABC]=2R2sinAsinBsinC
and we can calculate these values using law of cosines and then using sin2ΞΈ+cos2ΞΈ=1
We get that the area of MAβMBβMCβ is 7307ββ and thus [AEF]=41β[MAβMBβMCβ]=14157βββ.
OLD SOLUTION
Let x=β BAD=β CAD,y=β CBE=β ABE, and z=β BCF=β ACF. Notice that x+y+z=90β. In β³ABD, segment BE is the bisector of β ABD, and E lies on the perpendicular bisector of side AD. Therefore E is the midpoint of arc AD on the circumcircle of β³ABD. It follows that β BED=β BAD=x and β EDA=β EBA=y. Likewise, ACDF is cyclic, β CFD=β CAD=x, and β FDA=β FCA=z. Because EF is the perpendicular bisector of AD, triangles AEF and DEF are congruent, implying that
