always_correct wrote:

Here's a problem of mine!

Denote by $p_{XYZ}$ the perimeter of $\triangle XYZ$ .
Let $ABC$ be a triangle with point $D$ lying on $BC$ beyond $B$ such that $2 \cdot DC = p_{ABC}$ . Collinear points $E,P,Q$ are situated such that $E$ is the midpoint of $\overline{BC}$ , $ED = EP$ , and $EA = EQ$ . Points $R,S$ are the projections from $P$ to $\overline{BC}$ and $Q$ to $\overline{PR}$ , respectively. If $\angle ABC = \angle ACB < 45^\circ$ , prove that $p_{ABC} = p_{SBC}$ .

$[asy] size(10cm); pair A = (0, 0.5), B = (-1, 0), C = (1, 0), D = (-1.118, 0), E = origin, P = (0.52, 0.99), Q = (0.233, 0.443); pair R = (0.52, 0), S = (0.52, 0.443); draw(A--B--C--cycle); draw(D--E--P--cycle); draw(E--A); draw(P--R); draw(Q--S); draw(B--S--C); draw(rightanglemark(P,R,B,1.5)); draw(rightanglemark(P,S,Q,1.5)); label("$A$", A, N); label("$B$", B - (0, 0.05)); label("$C$", C - (0, 0.05)); label("$D$", D - (0, 0.05)); label("$E$", E - (0, 0.05)); label("$P$", P, N); label("$Q$", Q, - (0.05, -.05)); label("$R$", R - (0, 0.05)); label("$S$", S + (0.05, 0)); [/asy]$

It has a very cool solution! (I hope)

Hmm, it seems after a day nobody has a solution for me...

Time to post my own!

$[asy] size(10cm); pair A = (0, 0.5), B = (-1, 0), C = (1, 0), D = (-1.118, 0), E = origin, P = (0.52, 0.99), Q = (0.233, 0.443), Qp = (0.233, 0); pair R = (0.52, 0), S = (0.52, 0.443); draw(A--B--C--cycle); draw(D--E--P--cycle); draw(E--A); draw(P--R); draw(Q--S); draw(B--S--C); draw(Q--Qp); draw(rightanglemark(P,R,B,1.5)); draw(rightanglemark(P,S,Q,1.5)); draw(rightanglemark(Q,Qp,E,1.5)); draw(ellipse(E, 1.118, 0.5), blue+linewidth(1)); draw(circle(E, 0.5), red+linewidth(1)); draw(circle(E, 1.118), green+linewidth(1)); label(scale(0.7)*"$A$", A, N); label(scale(0.7)*"$B$", B - (0, 0.05)); label(scale(0.7)*"$C$", C - (0, 0.05)); label(scale(0.7)*"$D$", D - (0.05, 0.04)); label(scale(0.7)*"$E$", E - (0, 0.05)); label(scale(0.7)*"$P$", P, N); label(scale(0.7)*"$Q$", Q + (0, 0.09)); label(scale(0.7)*"$R$", R - (0, 0.05)); label(scale(0.7)*"$S$", S + (0.05, 0.02)); label(scale(0.7)*"$Q'$", Qp - (0, 0.05)); [/asy]$

Firstly note because $DB + DC = AB + AC$ , that $D, A$ lie on an ellipse with foci $B, C$ .

Let the blue ellipse with foci $B, C$ containing $A$ be $\mathcal{B}$ , the red circle with center $E$ containing $A$ be $\mathcal{R}$ , and the green circle with center $E$ containing $D$ be $\mathcal{G}$ . Finally, let $Q'$ be the projection from $Q$ to $\overline{BC}$ .

Consider the horizontal stretching $\mathcal{S}$ that takes $\mathcal{R} \to \mathcal{B}$ . It is equivalent to a homothety $\mathcal{H}$ at $E$ taking $\mathcal{R} \to \mathcal{G}$ followed by a vertical stretch $\mathcal{V}$ (as $\angle A$ is obtuse) with factor $k =\frac{R}{b}$ where $R$ is the radius of $\mathcal{G}$ and $b$ is the semi-minor axis of $\mathcal{B}$ . In other words,
$\mathcal{S} = \mathcal{V} \circ \mathcal{H}$
As $\triangle ABC$ is isosceles,
$k = \frac{R}{b} = \frac{ED}{EA}$ Now, as $P \in \mathcal{G}$ and $Q \in \mathcal{R}$ , it follows $k = \frac{EQ}{EP} = \frac{RS}{RP}$ by similarity of $\triangle EQQ'$ and $\triangle EPR$ .

Finally, note that $\mathcal{H}(Q) = P$ and as $k = \frac{RS}{RP}$ , $\mathcal{V}(Q) = S$ . Thus
$\mathcal{S}(Q) = \mathcal{V}(\mathcal{H}(Q)) = S$
As $\mathcal{S}(\mathcal{R}) = \mathcal{B}$ it follows $S \in \mathcal{B}$ , and as $B,C$ are the foci of $\mathcal{B}$ , we end up with
$AB + AC = SB +SC$ and we're done!

Mathematical