Olympiad Geometry : Viviani, Ptolemy and INMO 1995

Recall the easily provable Viviani's theorem: The sum of perpendicular lengths drawn from any point within an equilateral triangle to its sides is invariant and equal to the altitude of the triangle.

A slight generalization of the $INMO$ $1995$ result is very often used in many Olympiad geometry problems around the world. To quote:

The reflection of the orthocentre $H$ across the midpoint $M$ of a side $BC$ of any triangle $ABC$ furnishes a point $T$ which is the point diametrically opposite to $A$ on the circumcircle.

I leave the proof to you, it is simple angle-chasing. Although, the proof of the converse (which is almost the same proof actually !) can be found in the solution to this another problem by me.

The above result powerfully tells us that since $TA$ is diameter of circumcircle of $\Delta AQR$ , $\measuredangle TQA =90$ and also $\measuredangle TRA =90$ .

In other words, $P$ , $Q$ and $R$ are feet of perpendiculars from $T$ to the sides of equilateral triangle $ABC$ . So, applying the Viviani's theorem we have:

$PT + QT + RT = h = \frac{\sqrt{3}}{2} \times AB$ ...(*)

Now, observe that quadrilateral $TRBP$ is cyclic.( $\measuredangle TRB =\measuredangle TPB =90$ ). So, $\measuredangle RTP =120$ . So, even quadrilateral $TRB'P$ will be cyclic. (Opposite angles will sum to $60 + 120 = 180$ ). So, apply Ptolemy's theorem on the quadrilateral $TRB'P$ , denoting $B'R=B'P=PR=a$ :

$B'T \times a = RT\times a + TP\times a \implies B'T = RT +TP$ .

Similiarly, we obtain:

$C'T = PT + TQ$

$A'T = QT + TR$

And adding,

$A'T + B'T + C'T = 2\times(RT + PT + QT) = \sqrt{3}\times AB = 3.$

(From the result (*) above and since $AB = \sqrt{3}$ )