Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Sunday, March 13, 2016

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

Labels:
altitude,
circle,
equilateral,
incircle,
inradius,
metric relations,
square,
triangle

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

ReplyDeleteSo 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

Dear Sumith,

DeletePlease review the algebra simplification

Thanks

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

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

Thanks Antonio

배덕락(Bae deokrak)

ReplyDeleteWe 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.