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- #1

- Jan 26, 2012

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You are given the lengths AB, AC and BC of the triangle, as well as the distances of all three of the vertices to the triangle's circumcenter O. The inner circle is tangent to (AB). Find the shaded area.

Solution (don't click if you want to work it out yourself!):

$$|OA|^2 = |OM|^2 + |AM|^2 = |OM|^2 + \left ( \frac{1}{2} |AB| \right )^2$$

Because the inner circle is tangent to (AB) at M (as $|OA| = |OB|$) its area is:

$$A_\text{inner} = \pi |OM|^2 = \pi \left [ |OA|^2 - \left ( \frac{1}{2} |AB| \right )^2 \right ]$$

And the area of the outer circle is just:

$$A_\text{outer} = \pi |OA|^2$$

Thus the area of the shaded region is:

$$A_\text{shaded} = A_\text{outer} - A_\text{inner} = \pi |OA|^2 - \pi \left [ |OA|^2 - \left ( \frac{1}{2} |AB| \right )^2 \right ] = \pi \left ( \frac{1}{2} |AB|\right )^2 = \frac{\pi}{4} |AB|^2$$

The maximum value for $|AB|$ is clearly $2 |OA|$ when [AB] is a diameter of the circumcircle, thus $A_\text{shaded} \leqslant A_\text{outer}$.