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Source: Ukraine journal

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#1 h Jul 10, 2008, 6:37 PM • 2 Y

Y by Adventure10, Mango247

Let $I$ be the incenter of triangle $ABC$ . The circumcircles of triangles $AIC,AIB$ intersect sides $AB,AC$ at $K,N$ respectively. Let $M$ be an arbitrary point on segment $KN$ . Prove that the sum of the distances from $M$ to the sides of triangle $ABC$ does not depend on the choice of $M$ .

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#2 h Jul 12, 2008, 9:44 AM • 2 Y

Y by Adventure10, Mango247

Are you sure that this is the statement from the journal? I see two values for $M \equiv K$ and $M \equiv N$ .

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#3 h Jul 13, 2008, 1:31 AM • 2 Y

Y by Adventure10, Mango247

Most likely, $\triangle ABC$ is not general, but such that circles $\odot(AIC), \odot(AIB)$ intersect side segments $AB, AC$ (not just sidelines) again at $K, N.$ However, the proposition holds for a general $\triangle ABC$ and for the line $KN$ (not just segment), if the distances of $M$ from $BC, CA, AB$ are properly directed: Let $D, E, F$ be feet of perpendiculars from $M \in KN$ to $BC, CA, AB.$ Let $\overline{MD} > 0$ when $M$ is on the same side of $BC$ as $A$ and $\overline{MD} < 0,$ when $M$ is on the opposite side of $BC$ than $A.$ $\overline{ME}, \overline{MF}$ are directed similarly.
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Since $\odot(AIC), \odot(AIB)$ are centered on the internal bisectors of the angles $\angle B, \angle C,$ $\overline {BK} = \overline{BC}$ and $\overline {CN} = \overline{CB}.$ Then $\overline{AK} = c - a$ and $\overline {AN} = b - a.$ ( $K, N$ are on the segments $AB, AC$ if $a = BC = BK < BA = c$ and $a = CB = CN < CA = b,$ $a$ is the shortest triangle side.) Let $X \in BC, Y \in CA, Z \in AB$ be feet of the external bisectors of the $\angle A, \angle B, \angle C.$ From $\overline{AY} = \frac {bc}{c - a}, \overline {AZ} = \frac {bc}{b - a},$ we get $\frac {\overline {AN}}{\overline {AK}} = \frac {b - a}{c - a} = \frac {\overline {AY}}{\overline {AZ}},$ which means that the lines $KN \parallel XYZ$ are parallel.
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Let $P \in XYZ$ be arbitrary and $D, E, F$ feet of perpendiculars from $P$ to $BC, CA, AB.$ WLOG, assume $P$ is on the finite segment $YZ,$ so that $\overline {PE}, \overline {PF} > 0$ and $\overline {PD} < 0.$ (Proof for a segment of $XYZ$ containing the infinite point of this line is the same, except 2 directed distances are negative and one positive.) Let perpendicular to the internal bisector $AI$ of the angle $\angle A$ through $P$ cut $AC, AB$ at $P_b, P_c.$ $P_bP_c$ is locus of points $P',$ such that $\overline{P'E'} + \overline {P'F'} = \text{const}$ (where $E', F'$ are feet of perpendiculars from $P'$ to $AC, AB$ ). Let $F_b$ be foot of the perpendicular from $P_b$ to $AB,$ so that $\overline {PE} + \overline {PF} = \overline {P_bF_b}.$ The $\triangle P_bPY \sim \triangle AXY$ are centrally similar, having parallel sides, with similarity center $Y.$ Let a parallel to $BC$ through $P$ cut $AC, AB$ at $V, W$ and let a parallel to $AB$ through $P_b$ cut $BC$ at $U.$ Let the lines $VPW, P_bU$ intersect at $Q.$ The $\triangle P_bQV \sim \triangle ABC$ are again centrally similar with the same similarity center $Y,$ $Y \in BQ.$ Thus the external bisector $BYQ$ of the angle $\angle B$ is the diagonal of the parallelogram $UBWQ,$ it is a rhombus and

$\overline {PD} + \overline {PE} + \overline {PF} = \overline {PD} + \overline {P_bF_b} = 0.$

Let $I_a, I_b, I_c$ be the excenters of the original $\triangle ABC.$ Let an arbitrary parallel to $XYZ$ cut $BC, CA, AB$ at $X', Y', Z'.$ Let parallels to $AX, BY, CZ$ through $X', Y', Z'$ pairwise intersect at $I_a^*, I_b^*, I_c^*.$ The $\triangle I_a^*I_b^*I_c^* \sim \triangle I_aI_bI_c$ are centrally similar, having parallel sides. The $\triangle AY'Z' \sim \triangle AYZ$ are centrally similar with similarity center $A,$ hence $I_a^* \in AI_a$ and likewise, $I_b^* \in BI_b, I_c^* \in CI_c.$ Therefore, the incenter $I$ is the similarity center of the $\triangle I_a^*I_b^*I_c^* \sim \triangle I_aI_bI_c.$ Their orthic triangles $\triangle A^*B^*C^* \sim \triangle ABC$ are centrally similar with the same similarity center. The line $X'Y'Z'$ therefore cuts the lines $B^*C^* \parallel BC,$ $C^*A^* \parallel CA,$ $A^*B^* \parallel AB$ at the feet $X^*, Y^*, Z^*$ of the external bisectors of the $\angle A^*, \angle B^*, \angle C^*$ of the $\triangle A^*B^*C^*.$ Let $M \in X'Y'Z' \equiv X^*Y^*Z^*$ be arbitrary, let $D', E', F'$ be the feet of perpendiculars from $M$ to $BC, CA, AB$ and $D^*, E^*, F^*$ the feet of the same perpendiculars in $B^*C^*, C^*A^*, A^*B^*.$ Then

$\overline{MD'} + \overline{ME'} + \overline{MF'} =$

$= (\overline{MD^*} + \overline{D^*D'}) + (\overline{ME^*} + \overline{E^*E'}) + (\overline{MF^*} + \overline{F^*F'}) =$

$= \overline{D^*D'} + \overline{E^*E'} + \overline{F^*F'} = \text{const}$

which is a constant independent of $M \in X'Y'Z' \parallel XYZ.$
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The above result can be used in solving the following problems:
http://www.artofproblemsolving.com/viewtopic.php?t=209599
http://www.artofproblemsolving.com/viewtopic.php?t=209657

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