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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:$2-3-4-5-6$)

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peta arantes
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2 Answers2

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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 $QR$. enter image description here
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)

Anonymous
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If $O$ is circumcenter in interior of $\triangle ABC$, no integer value is possible for $OP + OQ + OR$.

We have a well known inequality that for any point $O$ interior to a triangle,

$OA + OB + OC \gt s$ where $s$ is sub-perimeter.

Here, $OA = OB = OC = R$ (circumradius)

So, $R \gt \dfrac43$ and also $R + r \gt \dfrac 43$

As the circumcenter $O$ is inside the triangle, we have an acute triangle and for an acute triangle, $2R + r \lt s\implies R + r \lt 4 - R \lt \dfrac 83$ (see more details at the end of the answer)

Now using Carnot's theorem as mentioned by you, $OP + OQ + OR = R + r$. So,

$ \displaystyle OP + OQ + OR \lt \frac83$

Edit: As corrected by Anonymous in comments, there is a stronger lower bound on $OP + OQ + OR$ than $4/3$ -

As $P, Q, R$ are midpoints of sides $AB, BC, AC$, using midpoint theorem and triangle inequality,

$OP + OQ \gt PQ = \frac{AC}{2}$

And that leads to $2 (OP + OQ + OR) \gt \frac{AB + BC + AC}{2} \implies OP + OQ + OR \gt 2$

As $2 \lt OP + OQ + OR \lt 8/3$, no integer value is possible.

you may follow this answer Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$ that shows the identity:

$ \displaystyle s^2 - (2R + r)^2 = 4 R^2 \cos A \cos B \cos C$

As $\cos A \cos B \cos C \gt 0$ for an acute triangle, it follows that $s \gt 2R + r$

Math Lover
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