# $3$ median of a triangle $ABC$ divided this triangle into $6$ part. Prove that circumcentre of each triangle is circumscribed.

$3$ median of a triangle $ABC$ divided this triangle into $6$ parts.

How do I prove that circumcentre of each triangle is circumscribed?

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I am sorry, but I am having a difficult time understanding your problem as posed. Can it be cleaned up? – Amzoti Dec 27 '12 at 3:54
@Haruboy15: Nice picture! I suggest you change the second sentence to "Prove that the circumcentre of each triangle is in the circumscibed circle of $\triangle ABC$." – André Nicolas Dec 27 '12 at 4:28
Join the circumcenters in an appropriate order ---say $KJHIMN(K)$--- to get a hexagon whose opposite edges are parallel. (Eg, $KJ \parallel IM$ because both are $\perp$ to $AD$.) Each pair of opposite edges, therefore, intersects at a point on the "line at infinity"; by a converse of a special case of Pascal's Theorem (en.wikipedia.org/wiki/Pascal%27s_theorem), the points at least lie on a conic. The article "Hexagons with opposite sides parallel" ( jstor.org/stable/3621413 ) relates such hexagons to triangle Cevians and discusses general conditions that force circularity. – Blue Dec 27 '12 at 6:22

Take the centroid ($O$) to be the origin, and place $A$ on the $x$-axis. With medians $AD$, $BE$, $CF$ (slightly different than the OP's diagram ($E$ and $F$ swapped)), we can write

$$A = -4(b+c,0) \qquad B = \phantom{-}4(b,k) \qquad C = \phantom{-}4(c,-k)$$ $$D = \phantom{-}2(b+c,0) \qquad E = -2(b,k) \qquad F = -2(c,k)$$

Consider four circumcenters associated with median $AD$. With "$XY^\perp$" abbreviating "perpendicular bisector of segment $XY$", we have that $OA^\perp$ (namely, $x=-2(b+c)$) meets $OE^\perp$ at a circumcenter $P$ with $y$-coordinate $(b^2+2bc-k^2)/k$; it meets $OF^\perp$ at circumcenter $Q$, with $y$-coordinate $-(c^2+2bc-k^2)$. The midpoint of vertical segment $PQ$ has $y$-coordinate $\frac{1}{2k}(b^2-c^2)$. This midpoint is the same as the midpoint of $RS$, where circumcenters $R$ and $S$ are the points at which $OD^\perp$ meets, respectively, $OB^\perp$ and $OC^\perp$.

Thus, $PQ$ and $RS$ have a common perpendicular bisector, so that $PQRS$ is an isosceles trapezoid, which is necessarily cyclic. That is, the four circumcenters associated with $AD$ lie on a common circle; likewise, those associated with $BE$, as well as those associated with $CF$. This (over-)accounts for all six circumcenters.

I'm thinking there should be an elegant, coordinate-free path to the statement "$PQ$ and $RS$ have a common perpendicular bisector".

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Without loss of generality, we can take the vertices of $\triangle ABC$ to be $A(6a,0),B(6b,0)$ and $C(0,6c)$ where $a<b$

So, the centroid $G(\frac{6a+6b+0}3,\frac{0+0+6c}3)$ or $(2(a+b),2c)$

The midpoint of $AB$ will be $D(\frac{6a+6b}2,\frac{0+0}2)$ or $(2(a+b),0)$

If the circum-centre of $\triangle DGB$ be $P(h,k)$

$$(h-6b)^2+(k-0)^2=\{h-3(a+b)\}^2+(k-0)^2=\{h-2(a+b)\}^2+(k-2c)^2$$

Solving the first two equation, $h=\frac{3a+9b}2$

Now, $h-6a=\frac{9(b-a)}2>0$ and $h-9b<0\implies 6b>h>6a$

Similarly, for $k$

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