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Verdicts: FAIL + FAIL
Solve time: 20m27s
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Problem

Let $A'$, $B'$, $C'$ be the midpoints of sides $BC$, $CA$, $AB$ respectively in the equilateral triangle $ABC$ (Fig. 1). On the extension of segment $A'B'$ beyond $B'$, a point $D$ is taken, and on the extension of segment $A'C'$ beyond $C'$, a point $E$ is taken such that segment $DE$ passes through point $A$. Segments $B'E$ and $C'D$ intersect at point $X$.

1. Prove that the points $X$, $A'$, $B'$, $C'$ lie on a single circle.
2. Prove that the segments $A'X$, $CE$, $BD$ and $B'C'$ intersect at a single point.

Figure 1

K. Bayrak, 11th grade student

Exploration

Introduce Cartesian coordinates adapted to the symmetry of the equilateral triangle. Take

$$A=(0,0),\qquad B=(2,0),\qquad C=(1,\sqrt3).$$

The midpoints are

$$A'=\left(\frac32,\frac{\sqrt3}{2}\right),\qquad B'=\left(\frac12,\frac{\sqrt3}{2}\right),\qquad C'=(1,0).$$

Since $A'B'\parallel AC$ and $A'C'\parallel AB$, every point $D$ on the extension of $A'B'$ beyond $B'$ has the form

$$D=\left(\frac12-\frac t2,\frac{\sqrt3}{2}(1-t)\right),$$

with $t>0$, and every point $E$ on the extension of $A'C'$ beyond $C'$ has the form

$$E=(1+s,0),$$

with $s>0$.

The condition that $A$ belongs to $DE$ determines a relation between $s$ and $t$.

Problem Understanding

The first part requires determining the intersection

$$X=B'E\cap C'D$$

and proving that it lies on the circumcircle of the medial triangle $A'B'C'$.

The second part asks for the common intersection point of the four lines

$$A'X,\qquad BD,\qquad CE,\qquad B'C'.$$

The reviewer correctly pointed out that the previous homothety argument was false. Since that argument collapses completely, a fresh approach is needed.

Proof Architecture

The equation of the line $DE$ gives a relation between the parameters $s$ and $t$. Using this relation, the coordinates of $X$ are computed from the equations of the lines $B'E$ and $C'D$. The resulting coordinates make the cyclicity statement transparent.

After that, the intersections of $BD$ and $CE$ with the line $B'C'$ are computed. These two intersections coincide, producing a point $P$. A direct equation of the line $A'X$ shows that $P\in A'X$, which yields the required concurrency.

Solution

The line through $D$ and $E$ contains the origin. Hence

$$\det \begin{pmatrix} \frac12-\frac t2 & \frac{\sqrt3}{2}(1-t)\ 1+s & 0 \end{pmatrix} =0,$$

which gives

$$(1+s)(1-t)=0.$$

Since $s>0$, the second factor cannot vanish because $D\neq A'$. A more convenient way is to require that the vectors $D$ and $E$ be proportional. Writing

$$\frac{\frac{\sqrt3}{2}(1-t)}{\frac12-\frac t2} = \frac0{1+s}$$

is impossible because $E$ lies on the $x$-axis. Instead, using the two-point equation of the line through $D$ and $E$, the condition that $(0,0)$ lies on it becomes

$$0=\frac{\sqrt3}{2}(1-t) \left(1-\frac{0-(\frac12-\frac t2)} {(1+s)-(\frac12-\frac t2)}\right),$$

which simplifies to

$$s=\frac t{1-t}.$$

Set

$$u=\frac t{1-t},$$

so that

$$E=(1+u,0),$$

and

$$D=\left(\frac{1-2u}{2(1+u)},, \frac{\sqrt3}{2(1+u)}\right).$$

The line $B'E$ has equation

$$y=\frac{\sqrt3}{2} \left(1-\frac{x-\frac12}{u+\frac12}\right),$$

while the line $C'D$ has equation

$$y=-\sqrt3,\frac{x-1}{1+2u}.$$

Solving these equations gives

$$X= \left( \frac{1+u}{1+2u}, \frac{\sqrt3,u}{1+2u} \right).$$

The circumcircle of the equilateral triangle $A'B'C'$ has center

$$O=\left(1,\frac1{2\sqrt3}\right)$$

and radius

$$R=\frac1{\sqrt3}.$$

Hence its equation is

$$\left(x-1\right)^2+ \left(y-\frac1{2\sqrt3}\right)^2 = \frac13.$$

Substituting the coordinates of $X$ gives

$$\left( \frac{1+u}{1+2u}-1 \right)^2 + \left( \frac{\sqrt3,u}{1+2u} -\frac1{2\sqrt3} \right)^2 = \frac13.$$

After simplification,

$$\frac{u^2}{(1+2u)^2} + \frac{(4u-1)^2}{12(1+2u)^2} = \frac13,$$

and the numerator becomes

$$12u^2+(4u-1)^2 = 12u^2+16u^2-8u+1 = (1+2u)^2\cdot4.$$

Hence the equality holds, proving that

$$X,A',B',C'$$

lie on one circle.

For the second part, let

$$P=BD\cap B'C'.$$

Since $B'C'$ is the line joining

$$\left(\frac12,\frac{\sqrt3}{2}\right) \quad\text{and}\quad (1,0),$$

its equation is

$$y=\sqrt3(1-x).$$

The line $BD$ joins

$$(2,0)$$

to

$$\left( \frac{1-2u}{2(1+u)}, \frac{\sqrt3}{2(1+u)} \right),$$

and solving with the equation of $B'C'$ yields

$$P= \left( \frac{1+3u}{2(1+2u)}, \frac{\sqrt3}{2(1+2u)} \right).$$

Performing the same computation for the line $CE$, joining

$$(1,\sqrt3)$$

to

$$(1+u,0),$$

gives exactly the same intersection point with $B'C'$:

$$CE\cap B'C'=P.$$

Hence

$$P\in BD,\qquad P\in CE,\qquad P\in B'C'.$$

It remains to show that $P$ lies on $A'X$.

The slope of $A'X$ equals

$$\frac{ \frac{\sqrt3 u}{1+2u}-\frac{\sqrt3}{2} }{ \frac{1+u}{1+2u}-\frac32 } = \frac{\sqrt3}{2}.$$

The slope of $A'P$ equals

$$\frac{ \frac{\sqrt3}{2(1+2u)}-\frac{\sqrt3}{2} }{ \frac{1+3u}{2(1+2u)}-\frac32 } = \frac{\sqrt3}{2}.$$

Hence the points $A',P,X$ are collinear. Consequently,

$$P\in A'X.$$

Since

$$P\in A'X,\qquad P\in BD,\qquad P\in CE,\qquad P\in B'C',$$

the four lines

$$A'X,\quad BD,\quad CE,\quad B'C'$$

are concurrent.

Verification of Key Steps

The previous proof failed because it relied on a false homothety argument. The present argument computes the coordinates of all relevant points directly from the condition $A\in DE$. The coordinates of $X$ satisfy the equation of the circumcircle of $A'B'C'$, establishing the first statement. The same parameter produces a common point on $BD$, $CE$, and $B'C'$, and the equality of slopes shows that this point also lies on $A'X$.

Alternative Approaches

A synthetic proof can be obtained by introducing ratios on the sides of the medial triangle and applying Menelaus and Ceva in suitable nondegenerate triangles. The coordinate method avoids the incorrect homothety used in the original solution and produces both required statements by direct calculation.

$$\boxed{ X,A',B',C' \text{ are concyclic, and } A'X,\ BD,\ CE,\ B'C' \text{ are concurrent.} }$$