Answer is $2h$ just reflect $H$ over $BC$ to get $H'$ and show that $AH'=BH'+CH'$.

Simple trig 2

Synthetic Vrangr wrote: Let $ABC$ be an acute-angled triangle with altitude $AT = h$. The line passing through its circumcenter $O$ and incenter $I$ meets the sides $AB$ and $AC$ at points $F$ and $N$, respectively. It is known that $BFNC$ is a cyclic quadrilateral. Find the sum of the distances from the orthocenter of $ABC$ to its vertices. Solution 13. The answer is $\boxed{2h}.$ Let $AI\cap BC=X$ and $D$ be the foot of perpendicular from $I$ to $BC.$ Let $r=ID$ and $R=AO$ be the inradius and circumradius respectively. Without loss of generality assume $\angle B\ge \angle C.$ It is given that $FN$ is antiparallel to $BC$ with respect to $\angle BAC,$ also $AO$ is the isogonal of $AT$ with respect to $\angle BAC.$ Hence $AO\perp IO.$ Note that, \begin{align*} \angle IAO&= 90^{\circ}-\angle C-\tfrac{\angle A}2=\tfrac{\angle B-\angle C}2,\\ \angle DIX&=\angle TAX=\tfrac{\angle A}2-(90^{\circ}-\angle B)=\tfrac{\angle B-\angle C}2=\angle IAO. \end{align*}Therefore, $\triangle AIO\sim\triangle IDX\sim\triangle ATX,$ so \[\tfrac rh=\tfrac{IX}{AX};\qquad\tfrac{R}{h}=\tfrac{AI}{AX},\]adding, $R+r=h.$ Denote the orthocenter by $H,$ thus the required sum is, \begin{align*} AH+BH+CH&=2(R+r)=2h &\blacksquare \end{align*}

Same as ayan.

Instantly solved by the fact the the locus of $X$ s.t. $d(X,AB)+d(X,BC)+d(X,CA)=const$ is a line perpendicular to $OI$.

Mindstormer wrote: Instantly solved by the fact the the locus of $X$ s.t. $d(X,AB)+d(X,BC)+d(X,CA)=const$ is a line perpendicular to $OI$. Ya using directed distances it's trivial Cause all those lines are parallel to trilinear polar of I and this is perp to IO

Lol, I just bashed it once I got the similar triangles.

Simple enough. Trigonomtery works fine.

lemma let H be the orthocenter of ABC,then$$ AH+BH+CH=2(R+r)$$where R is the circumradius of$$\triangle ABC$$and r the inradius proof Note that$$ r=4Rsin(A\div 2)sin(B\div 2)sin(C\div 2)$$, $$AH+BH+CH=2R(cosA+cosB+cosC)$$and $$4sin(A\div 2)sin(B\div 2)sin(C\div 2)+1=cosA+cosB+cosC$$We get the desired result using these three equations Back to the main problem Let be theD be the intersection of A-altitude with side BC of the given $$\triangle ABC$$Let AI intersect BC at E since BFNC is cyclic,and AO and AD are isogonal w.r.t. to$$\angle A$$, $$\triangle AOI\sim \triangle ADE$$Hence$$,AO\div AD=AI\div AE \Rightarrow R\div AD=b+c\div 2s =[r\times(b+c)]\div 2\times \triangle$$Since$$ R\div AD= Ra\div 2\times \triangle$$, $$Ra=r(b+c)\Rightarrow 2(R+r)a=AD\times a$$Thus distance from the orthocenter to the vertices is $$2\times AD$$=$$2h$$EDIT:I took D instead of T