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

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

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## Monday, February 9, 2015

### Geometry Problem 1081: Equilateral Triangle, Inscribed Circle, Inradius, Tangent Circles, Radius, Tangent Line, Sangaku Japanese Problem

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http://s14.postimg.org/7bjgzefe9/pro_1081.png

ReplyDeleteDraw lines and points per attached sketch

We have OB= 2.r , OE= r, DB= r

Triangle ODF is equilateral

Along OD we have OD= OB-DB => a+ b= 2r-2a => 3a+b= 2r …. (1)

Along OE we have OF=OE-FE => a+b=r-b => a+2b=r.. ( 2)

Along OF we have OF=2b+4c=r ……(3)

From (1) and ( 2) we have a= 3/5 r and b= r/5

Replace value of b in (3) we get c= r/10

Consider the altitude of the triangle, we have

ReplyDelete3a + 4b + 4c = 3r

Consider thr radius of the large circle C1,

2a + b = 5b + 4c (=R)

a − 2b − 2c = 0

Consider the radius of the in-circle,

3b + 4c = r

Summarizing, we have

3a + 4b + 4c = 3r

a − 2b − 2c = 0

3b + 4c = r

Hence,

a=3r/5, b=r/5, r/10.

Easy to observe that

ReplyDeleter = 3b + 4c,

b + 2a = R = r + 2b and thus r = 2a - b,

2r = b + 3a.

Add the last two equations:to get 3r = 5a or a = 3r/5

Next b = 2a - r = 6r/5 - r = r/5,

c = r/4 - 3b/4 = r/4 - 3r/20 = 2r/20 = r/10