DL=DK=X SE=ET=Y PQ=Z without loss of generic DP<PE PE/DE=PN/DK DP/DE=PM/SE DP/DE=(Z-X)/(Y-X) PE/DE=(Y-Z)/(Y-X) PN=X*(Y-Z)/(Y-X) PM=Y(Z-X)/(Y-X) PN+PM=(XY-XZ+YZ-XY)/(Y-X)=Z=PQ So PN+PM= PQ

Attachments:

$ In \ barycentric \ coordonate \ have: \ D(a;b;0), E(a;0;c) \Rightarrow DE \ equation \ is: \\ \frac{x}{a}=\frac{y}{b}+\frac{z}{c} \Leftrightarrow \frac{S_{PBC}}{a}=\frac{S_{PAC}}{b}+\frac{S_{PAC}}{c} \Leftrightarrow |PQ|=|PM|+|PN| \ (if \ P\in[DE]).$

Dear Mathlinkers, I want to give a synthetic response to the problem. The notations are the same as these of "Plane Geometry". 2.PQ = LD + TE 2.PQ = DK + ES 2.PQ = 2.PN + 2.PM and we are done Sincerely Jean-Louis

$ In \ my \ solution \ |PQ|=|PM|+|PN| \ is \ true \ (\forall) P\in [DE].$

Construct K,L,M,J be projection of D,E on side ABC. PH//KE and PG//DJ and P is midpoint then 2(PG+PH)=DJ+KE. DEML is an trapezoid so midline PF=1/2(DL+EM). BE,CD is bisector then EK=EM, DJ=DL. From all above, we have PF=PH+PG

Define $f(P)=|PM|+|PN|-|PQ|$. This is linear function $g(x_P,y_P)$ in $x_P$ and in $y_P$ as distance of point to a line is such. So it is sufficient to prove that for two distinct points $P$ we have $f(P)=0$. Of course $f(D)=f(E)=0$ QED.

jayme wrote: The notations are the same as these of "Plane Geometry". 2.PQ = LD + TE I guess this notation is different than in post #2. Otherwise the equality is true iff $DL=ET\vee DP=PE$

mihai miculita wrote: $ In \ my \ solution \ |PQ|=|PM|+|PN| \ is \ true \ (\forall) P\in [DE].$ Ummm... I am pretty sure this cannot be true? The lengths are clearly monotonic whatever direction you move in(and its not hard to use similar triangles to literally find the length) Always, its a nice problem Let the feet of the altitudes from $D,E$ to $BC$ be $G_1, G_2$ and the foot of perpendicular from $D$ to $AC$ be $G_3$ and the foot of the perpendicular from $E$ to $AB$ be $G_4$. Notice that because $BE$ and $CD$ are angle bisectors of $\angle B$ and $\angle C,$ respectively, we have that $\triangle CG_3D\cong \triangle CG_1D$ and similarly for the other, so this gives that $DG_1=DG_3$ and $EG_4=EG_2$. From here, it is trivial to finish, just note that $PQ=\frac{G_4E+G_3D}{2},$ which gives the desired. Note: Although the idea was simple, I felt this problem was pretty difficult because it is not at all obvious to draw the altitudes.