Mystic cevians

It will be recalled that the length $AH$ is twice the perpendicular distance of $O$ from $BC$ . This elementary result can be proved by the similiarity of triangles $AHG$ and $DOG$ noting the Euler line and that the median is split in a $2:1$ ratio by $G$ . (Here, $G$ is the centroid and $D$ the midpoint of $BC$ .)
Suppose $K,L,M$ be the feet of altitudes from $A,B,C$ respectively. So, the circle $\gamma$ has $AH$ for diameter. This implies that it is the circle circumscribing the quadrilateral $ALHM$ , because $AH$ subtends a right angle at the point of intersection of $\gamma$ and $AC$ ( other than $A$ ) and that makes it coincident with $L$ . A similiar argument for $M$ .

Now use the secant theorem as:

$BM \times BA =BT^2$

But by the cyclic quadrilateral $MACK$ , again the secant theorem yields: $BM \times BA = BK \times BC$

We conclude: $BT^2 = BK \times BC$

By a similiar argument: $CU^2 = CK \times CB$

An intuitively obvious addition step yields: $BT^2 + CU^2 = BC^2 = 2014^2$

Now then, $BT-CU-2014$ form a pythagorean triplet with $2014$ as the largest member. A simple way to work out the $BT$ and $CU$ is to factorize $2014$ and hunt a triplet with a factor as the largest in the triplet viz. hunting for $x-y-19$ or $x-y-2$ or $x-y-53$ . The latter works : $28-45-53$ . Magnify by $38$ to $1064-1710-2014$ .

It so happens that this is the only triplet.(Why?)

The answer : $1064+1710 = 2774.$

Hope, that was exhaustive. And error free!