For reference:
In a triangle $ABC$, $O$ is an interior point. The perpendiculars $OP, OQ$ and $OR$ are drawn to the sides $AB, BC$ and $AC$ respectively. Calculate: $OP+OQ+OR$, knowing that these values are integers and that $P, Q$ and $R$ are midpoints, and the perimeter of the triangle ABC is $8$.
My progress:
Point O will be the circumcenter
Carnot's Theorem: In any acute triangle, the sum of the
distances from the circumcenter on each side of the triangle
is equal to the sum of the circumradius (R) with the inradius (r)
$OP+OQ+OR = R+r$
We know that:($r_a, r_b, r_c$ radius of ex-inscribed circumference)
$ab+ac+bc=p^{2}+r^{2}+4Rr\\
OP = \frac{2R+r-r_b}{2}\\
OQ = \frac{2R+r-r_a}{2}\\
OR = \frac{2R+r-r_c}{2}\\
\therefore OP+OQ+OR = \frac{6R+3r-(r_a+r_b+r_c)}{2}=\frac{3}{2}(2R+r)-\frac{1}{2}(r_a+r_b+r_c)$
but i am not able to proceed……???
(Answer alternatives:$2-3-4-5-6$)
Best Answer
Consider $OP + OQ > PQ, OQ + OR > QR$ and $OR + OP > RP$,
we have $OP + OQ + OR > \frac{PQ + QR + RP}{2} = \frac{AB + BC + CA}{4} = \frac{8}{4} = 2.$
On the other hand, let the height of $\triangle PQR$ be $PP'$, where $P'$ is on $QR$.
In $\triangle PP'Q$, since $\angle PP'Q = 90^\circ$ is the greatest angle, $PQ$ is the longest side.
Also, $OQ$$\perp$$CB$ and $OR$$\perp$$BA$
$\Rightarrow CQ=QB$ and $BR=RA$
$\Rightarrow RQ//AC$
$\Rightarrow POP'$ is a straight line.
Then, $OP + OQ < OP + OP' + P'Q < PQ + P'Q$ and $OP + OR < RP + P'R$.
Combining the previous two inequalities, $2OP + OQ + OR < PQ + QR + RP = 4$.
Similarly, $OP + 2OQ + OR < 4$ and $OP + OQ + 2OR < 4$.
Combining the previous three inequalities, $OP + OQ + OR < \frac{4 + 4 + 4}{4} = 3$.
Therefore, $2 < OP + OQ + OR < 3$, which implies no solutions.
Remark:
There is a stronger upper bound. (see Math Lover's answer)