Apply Hook Length Formula

Geometry Level pending

The square $BCDE$ is inscribed in circle $\omega$ with center $O$ . Point $A$ is the reflection of $O$ over $B$ . A "hook" is drawn consisting of segment $AB$ and the major arc $\widehat{BE}$ of $\omega$ (passing through $C$ and $D$ ). Assume $BCDE$ has area $200$ . To the nearest integer, what is the length of the hook?

The answer is 57.

1 solution

Sharky Kesa
Aug 20, 2017

Since the area of $BCDE$ is $200$ , the length $BC=10 \sqrt{2}$ . Thus, since $OB=OC$ , and $OB \perp OC$ , we have $OB=10$ by Pythagoras' Theorem in $OBC$ . Therefore, $AB=10$ .

Furthermore, the circumference of $\omega$ is $2 \times 10 \times \pi = 20 \pi$ . Thus, the length of major arc $\widehat{BE}=\frac{3}{4} \times 20 \pi = 15 \pi$ .

Overall, we get the total length of the hook to be $10+15 \pi \approx 57$ .

Apply Hook Length Formula

The answer is 57.

1 solution

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