Drawing Circles Isn't Enough!

Firstly, we will prove that this shape is that of a circle. Firstly, for any point inside the circle, we will prove that the resultant locus is an ellipse.

Let the centre of the circle be $O$ , with radius $R$ , and the point on the interior be $X$ . The distance between any point $Q$ inside and the circumference of the circle is simply $R-OQ$ . Thus, we have $R-OQ=XQ$ , so $XQ+OQ=R$ . Therefore, the locus of $Q$ is an ellipse with a semi-major axis of $\frac {1}{2} R$ .

Note that $M$ passes through the midpoints of $AH$ , $BH$ and $CH$ , since $A$ , $B$ and $C$ are points on the circumcircle. We will now prove $M$ passes through $D$ , $E$ and $F$ , the feet of the altitudes from $A$ , $B$ and $C$ .

We will prove that the reflection of $H$ over each side lies on the circumcircle. Let $H'$ be the intersection of $AD$ and the circumcircle. Since $\angle HDB = 90^{\circ}$ , $\angle H'DB = 90^{\circ}$ . Since $AEDB$ is cyclic ( $\angle AEB = \angle ADB = 90^{\circ}$ ), $\angle EAD = \angle EBD = \angle HBD$ . Since $ACH'B$ is cyclic, $\angle DAE = \angle H'AC = \angle H'BC = \angle H'BD$ . Therefore, $\angle H'BD = \angle HBD$ . Since $\Delta HBD$ and $\Delta H'BD$ share $BD$ , $\Delta HBD \cong \Delta H'BD$ . Therefore, $HD=H'D$ , so $H'$ is the reflection of $H$ on $BC$ . Therefore, $M$ must pass through $D$ . We can prove similarly for $E$ and $F$ .

Thus, our locus contains $D$ , $E$ and $F$ as well as the midpoints of $AH$ , $BH$ and $CH$ . Hang on, this setup looks familiar. Indeed, it is the awesome Nine-Point Circle . A proof of this is left as an exercise to the reader, but since we get a circle, we must have the semi-major axis being equal to half the radius of this circle. Therefore, the radius of the circle traced by $M$ is $2$ , so the area is $4 \pi = 12.57$ .

Let complex numbers a, b, c denote the vertices of triangle ABC. Let origin be placed at circumcenter, so that we get complex number associated with orthocenter as a+b+c. Let any point P on circumcircle be $Rcis(x)$ Then midpoint of PH, M will be given by $1/2$ $(a+b+c)$ $+$ $Rcis(x)$ $/$ $2$ .So, if the nine point centre of triangle ABC is denoted by n , then we get $m-n$ $=$ $Rcis(x)$ $/$ $2$ .Hence, taking modulus on both sides we get that $|m-n|$ $=$ $R/2$ .Hence the locus of M is the nine point circle of triangle ABC.

This locus is the circumcircle of the orthic triangle with radius r. But 2r=R=4, thus r=2.
$Required\ area = \pi r^2=4\pi=12.57.$
OR
We know that three feet of the altitudes, three midpoints on H to a vertex, all six points where M moves, this is likely to be the NinePoint Circle.
So 2r=R, as above.

if the question applies to all cases of triangles then it should satisfy the case of equilateral triangle ........... ..........in equilateral triangle circumcentre and orthocentre lie on same point that is on the centroid............... ........... therefore the locus of point M becomes circle of radius 2 units............. ...........so the area is 4*3.14 = 12.56