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

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

Details: Click on the figure below.

## Sunday, May 12, 2019

### Geometry Problem 1433: Tangent Circles, Diameter, Perpendicular, Midpoint, Measurement, Poster

Labels:
circle,
diameter,
geometric art,
measurement,
midpoint,
perpendicular,
perpendicular bisector,
poster,
Problem,
tangent,
typography

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https://photos.app.goo.gl/WkovAxHYaaVhSUNF8

ReplyDeleteconnect QF , TB, TA

Let QF meet circle Q at H

Since circle O tangent to circle Q at T so O, Q,T are collinear

Triangles OBT and QFT are isosceles with ∠ (BOT)= ∠ (FQT)

So ∠ (OTB)= ∠ (QTF) => B, F, T are collinear

In the same way we also have A,H,T are collinear

Triangle BEF and BTA are similar ( case AA)

So AT=EF. (AB/BF)….(1)

Triangle BEF similar to HKA ( case AA)

So AH=BF .(KH/BE)=BF.(EF/BE)…(2)

Multiply (1) by (2) => AT. AH= AG^2= EF.(AB/BF). (BF/BE).EF

AG^2= (AB/BE). EF^2= 4 EF^2

So AG= 2 EF

OCBD being a rhombus, Tr. ACD is equilateral

ReplyDeleteHowever circle Q is not unique.

Hence Peter's proof above does not/ need not use this fact

the fact that E is the midpoint of OD ( or AB/BE=4) is implicitly indicate that ACD is equilateral.

ReplyDeleteIn other word , if ACD is not equilateral , AG will not equal 2.EF

Peter Tran