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

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

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## Sunday, May 5, 2019

### Geometry Problem 1432: Tangent Circles, Secant, Tangent Lines, Proportionality, Similarity

Labels:
circle,
geometry problem,
proportionality,
proportions,
secant,
similarity,
tangent

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https://photos.app.goo.gl/uWKn5PxzEsWQnLBy9

ReplyDeleteLet R=TO1 and r=TO2

Let 2 α= angle AO1T= angle TO2B

In triangle AO1O2, using cosine formula we have

AO2^2= (R+r)^2+R^2-2R(R+r)cos(2 α)

And AE^2=a ^2=AO2^2-r^2

AE^2=2R(R+r)(1-cos2 α)

Replace 1-cos(2 α)=2 sin(α)^2 in above we get

AE^2= 4R(R+r)sin(α)^2 => a= 2.sqrt(R(R+r)).sin(α)

In the same way we have BG= b= 2.sqrt(r(R+r)).sin(α)

So a/b= sqrt(R/r)

And ratio of tangent from A to circle O2 and B to circle O1 is not depend on angle α

So c/d= sqrt(R/r) and a/b=c/d or a/c=b/d

connect AC, BD;

ReplyDeleteThrough T, make common tangent line.

It is easy to see triangle ACT similar to triangle BDT;

so AT/CT = BT/DT ;

ALSO a^2 = AT*AB; c^2 = CT*CD;

b^2 = BT*AB; d^2 = DT*CD;

SO a^2/c^2 = b^2/d^2

so a/c= b/d

By drawing the common tangent at T we can easily deduce that AC//BD

ReplyDeleteC^2/d^2 = CT.CD/DT.CD = CT/DT = AT/BT (because AC//BD) = AT.AB/BT.AB = a^2/b^2

From which a/c = b/d

Sumith Peiris

Moratuwa

Sri Lanka