complex bash

DottedCaculator wrote: complex bash sure, why not. Let $BCDE$ be inscribed in the unit circle, and let line $AC$ meet the unit circle again at $X$. Let the internal bisector of $\angle DCA$ meet the unit circle again at $Y$, so that $A$ is the reflection of $D$ over $CY$. Then $$|b|=|c|=|d|=|e|=|x|=|y|=1$$Since $BD = CD$, we have $$d^2 = bc$$Since $\angle DCA = \angle EBD$ as directed angles, we have $$\frac{d}x = \frac{e}b$$Since $Y$ is the midpoint of arc $DX$, we have $$y^2 = dx$$Thus we can reduce to three variables, writing \begin{align*} |b|=|d|=|y|&=1 \\ c &= \frac{d^2}b \\ x &= \frac{y^2}d \\ e = \frac{bd}x &= \frac{bd^2}{y^2} \\ a = c + y - cy\overline{d} &= \frac{by+d^2-dy}b \end{align*}Then we have $\angle BAC + \angle AED = 180^{\circ}$ as directed angles. Thus $$\left(\frac{b-a}{c-a}\right)\left(\frac{a-e}{d-e}\right) \in \mathbb{R}$$$$\frac{\frac{b^2-by-d^2+dy}b}{\frac{dy-by}b}\cdot\frac{\frac{by^3+d^2y^2-dy^3-b^2d^2}{by^2}}{\frac{dy^2-bd^2}{y^2}} \in \mathbb{R}$$$$\frac{(b+d-y)(by^3+d^2y^2-dy^3-b^2d^2)}{bdy(bd-y^2)} = \frac{(by+dy-bd)(y^3+b^2d-b^2y-bd^2)}{bdy(bd-y^2)}$$$$(b+d-y)(by^3+d^2y^2-dy^3-b^2d^2) = (by+dy-bd)(y^3+b^2d-b^2y-bd^2)$$Expanding out gives $$b^2y^3+bd^2y^2-bdy^3-b^3d^2+bdy^3+d^3y^2-d^2y^3-b^2d^3-by^4-d^2y^3+dy^4+b^2d^2y$$$$= by^4+b^3dy-b^3y^2-b^2d^2y+dy^4+b^2d^2y-b^2dy^2-bd^3y-bdy^3-b^3d^2+b^3dy+b^2d^3$$which we condense to $$2b^3dy-b^3y^2+2b^2d^3-b^2d^2y-b^2dy^2-b^2y^3-bd^3y-bd^2y^2-bdy^3+2by^4-d^3y^2+2d^2y^3 = 0$$Dividing out $b-y$ (which is nonzero since $B, Y$ lie on opposite sides of $AC$) gives $$2b^2dy-b^2y^2+2bd^3-bd^2y+bdy^2-2by^3+d^3y-2d^2y^2 = 0$$We factor this as $$(by+d^2)(2bd-by+dy-2y^2) = 0$$So one of these two factors is zero. If the first factor is zero, then we have \begin{align*} b &= -\frac{d^2}y \\ a &= \frac{y^2}d \\ e &= -\frac{d^4}{y^3} \end{align*}and so $$|a-b| = |d-e| = |d^3+y^3|$$If instead the second factor is zero, we have \begin{align*} b &= \frac{y(2y-d)}{2d-y} \\ a &= \frac{2d^3-3d^2y+2y^3}{y(2y-d)} \\ e &= \frac{d^2(2y-d)}{y(2d-y)} \\ \end{align*}Then we have the vectors \begin{align*} a-b = \frac{4d^4-8d^3y+2d^2y^2+8dy^3-6y^4}{y(2d-y)(2y-d)} &= \frac{2(d-y)(d+y)(2d^2-4dy+3y^2)}{y(2d-y)(2y-d)} \\ a-e = \frac{3d^4-4d^3y-d^2y^2+4dy^3-2y^4}{y(2d-y)(2y-d)} &= \frac{(d-y)(d+y)(3d^2-4dy+2y^2)}{y(2d-y)(2y-d)} \end{align*}and thus $|a-b| = 2|a-e|$. $\blacksquare$

From xibo: Let $BCDE$ be unit circle, $d=1$, then $c=\overline{b}$. If $(d-c)=w(a-c)$, then $e=w^2b$, where $w$ is a unit complex number. Then, express the condition and the result in terms of $c$, $w$. The calculation is very short.

We have $\frac{(a-c)^2}{(d-c)^2}=\frac be$. If $\omega=\frac{a-c}{d-c}$, then $a=c+\omega(d-c)$ and $b=\omega^2e$. We also have $bc=d^2$, so $\omega^2ce=d^2$. Then, $\frac{c-a}{b-a}\frac{d-e}{a-e}$ is real, so $$\frac{\omega}{\omega^2e-c-\omega(d-c)}\frac1{c-e+\omega(d-c)}=\frac{bc}{(d(c-\omega^2e)-\omega e(c-d))(d(e-c)\omega+(c-d)e)}$$$$\omega(d(c-\omega^2e)-\omega e(c-d))(d(e-c)\omega+(c-d)e)=d^2(\omega^2e-c-\omega(d-c))(c-e+\omega(d-c))$$$$(d(c^2-d^2)\omega-d^2(c-d))(d(d^2\omega-c^2\omega^3)+(c-d)d^2)=d^2(d^2-c^2-\omega c(d-c))(c^2\omega^2-d^2+\omega^3 c(d-c))$$$$((c+d)\omega-d)(d^2\omega-c^2\omega^3+(c-d)d)=-(c+d-\omega c)(c^2\omega^2-d^2+\omega^3 c(d-c)).$$$$(c+d\omega)(d-c\omega)(2c\omega^2-c\omega+d\omega-2d)=0.$$ The first case becomes $a=-\omega c$ so $|a-b|=|b+\omega c|=|\omega e+c|=|d-e|$. The second case gives $d=c\omega$, which is easily seen to be impossible by the convex condition. Otherwise, $AB=2AE$ is equivalent to $|a-b|^2=4|a-e|^2$ or $$(a-b)(\overline a-1/b)=4(a-e)(\overline a-1/e)$$$$\omega^2(c^2+c\omega(d-c)-d^2)(d^2\omega+d(c-d)-\omega c^2)=4(c^2\omega^2+c\omega^3(d-c)-d^2)(d^2\omega+d(c-d)-c^2\omega^3)$$$$-(2c\omega^2-c\omega+d\omega-2d)(2c^3\omega^4-c^3\omega^3-c^3\omega^2-2c^2d\omega^4-2c^2d\omega^3+2c^2d\omega^2-c^2d\omega+cd^2\omega^3-2cd^2\omega^2+2cd^2\omega+2cd^2+d^3\omega^2+d^3\omega-2d^3)=0$$which is true. Therefore, either $AB=DE$ or $AB=2AE$.

This is a hard problem. The key is to reflect A across the perpendicular bisector of BC.

Doesn't exist a synthetical solution, elementary, without complex numbers?

Conjuring problem! We present a synthetic solution. We will need the following well known lemma. Lemma: Let $ABCD$ be a quadrilateral, so that $\measuredangle ABD + \measuredangle CAD + \measuredangle ADB + \measuredangle CBD = 180^{\circ}$, then either $ABCD$ is cyclic or $AB \cdot CD = AD \cdot BC$. Moving back to the problem.

. If $ABDE$ is cyclic, then $ABDE$ is isosceles trapezoid, and so $AB = DE$. Now, moving onto to the more interesting case, we have $AB \cdot DE = AE \cdot BD$, this implies that $A,D$ lies on the appolonian circle with ratio $\frac{AB}{AE}$. Let $O$ be the center of this appolonian circle, then notice that $CO$ is perpendicular bisector of $AD$, this implies that $2\measuredangle OCD = \measuredangle BDO - \measuredangle OBD$, observe that this uniquely determines $O$, so does it determine $A,E$, we redefine points now, define $E$ as the point on $(BCDE)$ so that $DE = \frac{BD}{2}$, define $O = DD \cap BE$, and define $A$ by reflection mentioned earlier, it is easy to see these points satisfy all the conditions prescribed in the problem, hence we're done.