## Wednesday, June 11, 2014

### Geometry Problem 1020: Circle, Tangent, Chord, Perpendicular, Geometric Mean

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

Click the figure below to see the complete problem 1020.

1. Note that ∠(C1DA1) supplement to ∠(C1BC)
∠(BDC) supplement to ∠(C1BC) => ∠(C1DA1)= ∠(BDC) => ∠(C1DB)= ∠(A1DC)
So we have c1/a1= BD/DC…… (1)
Similarly we also have ∠(BDA1)= ∠(CDB1) so a1/b1=BD/DC….(2)
From ( 1) and (2) we have c1/a1=a1/b1 => a1= sqrt(b1.c1)

2. Let A = x
Thus we get <CBC1 = <BCB1 = 90 + x/2
BA1DC1 is smiliar to CA1DB1 because of 3 equal angles (The 2 pairs of 90 as given)
From simple looking and proportion we get a1/c1 = b1/a1
Finally a1=GM(b1,c1)
Is that good enough solution??

1. Note that 2 quadrilaterals with 3 corresponding congruence angles are not necessary similar.
Peter Tran

3. Triangle C1BD similar to A1CD, and triangle A1BD similar to B1CD due to tangent angles.
From above two statements c1/a1=BD/CD, and a1/b1=BD/CD can be derived respectively.
Therefore c1/a1=a1/b1 or a1^2=b1*c1.

4. From similar triangles c1/a1 = BD /CD = a1/b1 and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

5. Note further that A1B1 is a tangent to circle A1BC1D and similarly A1C1 to A1CB1D