Problem 1

Let

ABCD

be a parallelogram with

AB=a

AD=1

, and

\angle BAD=\alpha

. If

\triangle ABD

is acute, prove that the four circles of radius

1

centered at

A,B,C,D

cover the parallelogram if and only if

a\le\cos\alpha+\sqrt3\sin\alpha

Step 3 of 6: Express the circumradius

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BD^2=1+a^2-2a\cos\alpha,\quad R=\frac{BD}{2\sin\alpha}

Detailed analysis

The cosine law in $\triangle ABD$ gives $BD^2=1+a^2-2a\cos\alpha$ . Since $BD$ is opposite angle $\alpha$ , the sine law gives $BD=2R\sin\alpha$ , hence the displayed formula for $R$ .