## Sunday, March 13, 2016

### Geometry Problem 1199: Equilateral Triangle, Square, Altitude, Circle, Incircle, Inradius, Metric Relations

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1199.

1. Let DF = p, GH =q and AG = m

So BD = sqrt3 -1 and from similar triangles p = 1 - sqrt(1/3)

and p/1 = (1-q)/q from which q = (6+sqrt3)/11 upon simplifying the algebra

m^2 = q^2 + 1 and so simplifying again m = (1/11) sqrt(149-10sqrt3)

x = S(AGC)/s = (1/2)q X 2/{(1/2) (2m+2)} = 2q / (2m+2)

Substituting for q and m and again simplifying the algebra

x = (5 - sqrt3)/{11 + sqrt(149-10sqrt3)}

I have not tried to simplify this unwieldy expression but it can be done by rationalizing the denominator.

Sumith Peiris
Moratuwa
Sri Lanka

1. Dear Sumith,
Thanks

2. From my proof above x = q/(1+m) which upon simplifying gives

(6+sqrt3)/(11+ sqrt(160+12sqrt3))

Thanks Antonio

2. 배덕락(Bae deokrak)

We see that two triangles BDF, CEF are 30-60-90 triangle and so
CE=1, EF=1/sqrt3, FC 2/sqrt3, DF=1-1/sqrt3 , BD=sqrt3 –1 and BF=2sqrt3 –1
Furthermore, since two triangles FDG and AHG are similar, we have
DF/DG=AM/(1-DG) <=> DG=(5-sqrt3)/11 <=> HG= 1-DG=(6+sqrt3)/11
The area of triangle ACG =[tri_ACG]= HG= x(1+GA) =x(1+sqrt{1+HG^2}), that is,
x=HG/(1+sqrt{1+HG^2}). where HG=(6+sqrt3)/11.