A, B, C AND D ARE FOUR CONSECUTIVE POINTS ON A CIRCLE, SUCH THAT A...

3. A, B, C and D are four consecutive points on a circle, such that AB = 1, BC = 2,

CD = 3 and ∠CDA=60°. Determine all possible lengths of DA.

【Solution】

Since ∠CDA=60°, AC is the second longest side in triangle ACD. If DA ≥ 3, then

AC ≥ 3=AB+BC, which is a contradiction. Hence DA<3. (5 points) Let E be the point

on CD such that DE=DA. Then ADE is an equilateral triangle. We have

∠ = ° − ∠ = ° = ° − ∠ = ∠

180 120 180

AEC AED EDA ABC

Suppose ∠ BCD=60°. Then ABCE is a parallelogram and DA=EA=BC=2. ( 5 points )

F

B

A B

A

D E C

If ∠ BCD ≠ 60°, extend CB to F such that BF=1. Then BAF is an equilateral triangle,

so that ∠ BFA=60°. Now CF=CB+BF=2+1=3=CD. Hence ∠ CFD= ∠ CDF, so that

∠ = ∠ − ° = ∠ − ° = ∠ . It follows that DA=FA=AB=1. ( 10

60 60

AFD CFD CDF ADF

points )

In summary, there are two possible lengths of DA, 2 or 1 . ( answer only 5 points )

ANS: 2 or 1