# Fun geometry problem

#### Bacterius

##### Well-known member
MHB Math Helper
Here's a nice geometry problem, not hard at all if you can see what's really going on.

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!):

Let M be the midpoint of [AB]. Then (OM) is perpendicular to (AB). We know |OA|, so by Pythagoras we have:

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

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 $A_\text{shaded} \leqslant A_\text{outer}$.

#### Albert

##### Well-known member
What is the meaning of "x" ,in your diagram ?

people will misunderstand that x is the center of the small

circle .

I think your solution is based on the assumption that the

two circles are concentric circles

(but your problem did not reveal this fact )

If they are not concentric circles,the solution will be much

more complicated,

(in this case point M is not on the small circle)

How do you think ?

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#### Bacterius

##### Well-known member
MHB Math Helper
What is the meaning of "x" ,in your diagram ?

people will misunderstand that x is the center of the small

circle .

I think your solution is based on the assumption that the

two circles are concentric circles

(but your problem did not reveal this fact )

If they are not concentric circles,the solution will be much

more complicated,

(in this case point M is not on the small circle)

How do you think ?
There is no "x", it's indeed just a cross indicating the centre O of the inner and outer circles. I thought that was clear enough (and the diagram is not mine). The two circles must be concentric, it follows from the fact that the outer circle is the circumcircle of the triangle (stated in the problem) and that the inner circle is clearly tangent to the midpoint of AB, by |OA| = |OB|.