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

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

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## Wednesday, March 25, 2015

### Problem 1103: Right Triangle, Incircle, Inscribed Circle, Radius, Geometric Mean, Sangaku, Japanese, Metric Relations

Labels:
circle,
geometric mean,
incircle,
inscribed,
Japan,
metric relations,
radius,
right triangle,
sangaku

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http://s25.postimg.org/815s5zg3j/pro_1103.png

ReplyDeleteDraw lines per sketch

We have ∠ (OED1)= ∠ (OEO2)= ∠ (OFO1)= ∠ (ODO2)= 45 degrees

And ∠ (O1FD)= ∠ (FDO2)= ∠ (O1EO2)= 90 degrees

Calculate O1O2^2= O1E^2+O2E^2= 2. r1^2+ 2.r2^2

O1O2^2= FD^2+(DO2-FO1)^2

Replace FD^2= 2.r^2 , DO2= r2.sqrt(2) , FO1= r1.sqrt(2) in above expression

We get r^2=2.r1.r2

A/2 + C/2 = Pi/4,

ReplyDelete1 + tan C/2 = 1 + tan ( Pi/4 - A/2) = 1 + [(1 - tan A/2) /(1 + tan A/2)] = 2/(1 + tan A/2)

Let L, M be the projections of O1, O2 on AC resp..

C, O2, O are collinear and r/r2 = EC/ MC = 1 + EM/MC = 1 + O2M/MC = 1 + tan C/2 = 2/(1 + tan A/2)

A, O1, O are collinear and r/r1 = AE/AL = 1 + LE/AL = 1 + r1/AL = (1 + tan A/2)

Hence (r/r2)(r/r1) = 2, r^2 = 2.r1.r2

FODB is a square of side r so easily

ReplyDeleter = (c+a-b)/2.......(1)

From similar Tr. s (r-r1)/r = r1/(c-r)......(2)

Eliminating r we have

r1 = (c+a-b)/(c+b-a)/4c = {c^2 - (a-b)^2}/4c

Simplifying using c^ 2 = b^ 2 - a^2 we have

r1 = a(b-a)/2c and similarly

r2 = c(b-c)/2a

So 2r1r2 = (b-a) (b-c)/2.....(3)

Now from (1),

r ^2 = (c+a-b)^2 / 4 which simplifies to (b-a)(b-c) /2 which from (3) is therefore 2r1r2

Sumith Peiris

Moratuwa

Sri Lanka