See complete Problem 83

Area of a Excircle Contact Triangle, Exradius, Circumradius. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 83

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ang(FDE)=B;ang(EDG)=C;ang(FDG)=B+C=180-A

ReplyDelete[EFG]=[FDE]+[EDG]-[FDG]

= 0.5r_a²sinB+0.5r_a²sinC-0.5r_a²sinA

=0.5r_a²(-sinA+sinB+sinC)

with the law of sines

-sinA+sinB+sinC=(-a+b+c)/(2R)=(p-a)R

we know that S=r_a(p-a)

S_a=r_aS/(2R)

.-.

I propose a solution to problem 83 without law of sines.

ReplyDeleteAs BFDE is cyclic, ang(EDF)=180ยบ - ang(CBF) = ang(B). So, the area of tr. EDF is S(EDF) = 0,5.ra*2.sen B. As S = 0,5.a.c.sen B, then S(EDF)/S = (ra*2)/(ac). Similarly, S(EDG)/S =(ra*2)/(ab) and S(FDG)/S =(ra*2)/(bc).

Therefore Sa/S = (ra*2)/(ac) + (ra*2)/(ab) -(ra*2)/(bc) =((b+c-a).ra*2)/(abc) (1).

From the known properties of Elementary Geometry we have S=0,5.(b+c-a).ra and S=(abc)/(4R). These two equalities lead to b+c-a = (abc)/(2.ra.R).

This expression in (1) gives us Sa/S = (ra*2)(abc)/(2.ra.R.abc) = ra/(2R).

Pure Geometry solution

ReplyDeleteLet h be the altitude from A of triangle ABC and let m be the altitude from F of triangle DEF.

Now BF = s-c, CG = s-b, AF = s.

So, S(ABC) = 2.S(ADF) – 2.S(BDF) – 2. S(CDG) = 2ra s – ra(s-c) - ra(s-b) = ra(s-a) ….(1)

Further, S(ABC) = ha/2 …(2) and

S(DEF) = ram/2 …(3)

From the 3 similar triangles each of which have < B and < 90, we have,

h/c = b/2R = m/ ra ….(4)

From (2) and (4), S(ABC) = abc/4R …(5)

From (3) and (4), S(DEF) = S(ABC). ra^2/ac …(6)

Writing similar expressions for S(DEG) and S(DFG) we have

S(EFG) = S(DEF) + S(DEG) – S(DFG) = S(ABC). ra^2 (1/ac + 1/ab – 1/bc)

Hence S(EFG) = S(ABC). ra^2 (b+c-a)/abc = S(ABC). ra^2 {2(s-a)}/abc

Therefore Sa/S(ABC) = S(EFG)/S(ABC) = { ra(s-a)}(2ra/abc) = S(ABC). (2ra/abc) =(abc/4R)(2ra/abc)

Finally Sa/S(ABC) = ra/2R

(Note - ra is r subscript a = ex-radius related to A)

Sumith Peiris

Moratuwa

Sri Lanka