## Sunday, April 28, 2013

### Problem 872: Quadrilateral, Triangle, Angle, 30, 90 degrees

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

Click the figure below to see the complete problem 872.

1. http://img267.imageshack.us/img267/4236/problem872.png
Draw circle center B radius BA.
Draw common tangent to these 2 circles.
This tangent will parallel to BO
Circle O cut AD at F and ∠ (AFB)= 30 degrees
Note that BD is angle bisector of ∠ (ADE) and DE ⊥CO => ∠(CBF)= 45
And ∠(CDF)= ∠(BFA)= 30 degrees so x=15 degrees.
Verify that ∠(FBD)=15 and ∠ (CBD)= 45-15=30 degrees= 2.x

2. My trigonometric solution suggests that x may be 15° or 30°. (In both cases, it is the same 30°-60°-90° triangle). If it were the latter, would Peter's solution hold?

3. Must mention here that if x=30° then /_BCD=90° which is contrary to the given condition & thus x=30° has to be rejected and, therefore, x=15°

4. Ajit

http://imageshack.us/a/img29/3016/problem8721.png
Yes, there is 2nd solution with x= 30 .
let 2 circles intersect at A and C' and AD cut circle O at D'
note that D'C' tangent to circle O and angle BC'A= 30 , angle AD'B=angle BD'C'=30 and angle C'BD'=60 .

Peter

5. Peter Tran:

Why does the common tangent pass through C? You are going by the figure you are building, not proving such a fact. If x=20 degrees, you will see why.....

Erina-NJ

6. Erina

C is any point on circle O such that angle(CBD)=2.angle (CDB). D is the intersection of horizontal line through A and the tangent line from C to circle B.
We have 2 solutions :
solution 1: C is located on the common tangent line of circles O and B by accident
solution 2: C is located on the intersection of circle O and circle B ( not on common tangent line)
Peter

7. We are sure that C is on the Circle O, but are not sure that CD is tangent with Circle B. This is what needs to be clarified. If you can prove this then your 2 solutions are clear.
1) C is not the same as E, x=15 degrees
2) C is the same as E, x=30 degrees

Erina-NJ