Problem 1965

For the equivalence with the collinearity condition, let $F$ denote the foot of the perpendicular from $P^{\prime}$ to $A B$, so that $F$ is the midpoint of $P P^{\prime}$. We have that $P$ lies on $C E$ if and only if $F$ lies on $M N$, which occurs if and only if we have the equality $\angle A F M = \angle B F N$ of signed angles modulo $\pi$. By concyclicity of $A P^{\prime} F M$ and $B F P^{\prime} N$, this is equivalent to $\angle A P^{\prime} M = \angle B P^{\prime} N$, which occurs if and only if $A P^{\prime} M$ and $B P^{\prime} N$ are directly similar.

For the other equivalence with the area condition, we have the equality of signed areas $\operatorname{area}(A B D) + \operatorname{area}(A B P) = \operatorname{area}\left(A P^{\prime} B D\right) = \operatorname{area}\left(A P^{\prime} D\right) + \operatorname{area}\left(B D P^{\prime}\right)$. Using the identity $\operatorname{area}(A D E) - \operatorname{area}\left(A P^{\prime} D\right) = \operatorname{area}(A D E) + \operatorname{area}\left(A D P^{\prime}\right) = 2 \operatorname{area}(A D M)$, and similarly for $B$, we find that the area condition is equivalent to the equality
$$\operatorname{area}(D A M) = \operatorname{area}(D B N)$$
Now note that $A$ and $B$ lie on the perpendicular bisectors of $P^{\prime} E$ and $P^{\prime} C$, respectively. If we write $G$ and $H$ for the feet of the perpendiculars from $D$ to these perpendicular bisectors respectively, then this area condition can be rewritten as
$$M A \cdot G D = N B \cdot H D$$
(In this condition, we interpret all lengths as signed lengths according to suitable conventions: for instance, we orient $P^{\prime} E$ from $P^{\prime}$ to $E$, orient the parallel line $D H$ in the same direction, and orient the perpendicular bisector of $P^{\prime} E$ at an angle $\pi / 2$ clockwise from the oriented segment $P^{\prime} E$ - we adopt the analogous conventions at $B$.)

To relate the signed lengths $G D$ and $H D$ to the triangles $A P^{\prime} M$ and $B P^{\prime} N$, we use the following calculation.
Claim. Let $\Gamma$ denote the circle centred on $D$ with both $E$ and $C$ on the circumference, and $h$ the power of $P^{\prime}$ with respect to $\Gamma$. Then we have the equality
$$G D \cdot P^{\prime} M = H D \cdot P^{\prime} N = \frac{1}{4} h \neq 0 .$$
Proof. Firstly, we have $h \neq 0$, since otherwise $P^{\prime}$ would lie on $\Gamma$, and hence the internal angle bisectors of $\angle E D P^{\prime}$ and $\angle P^{\prime} D C$ would pass through $A$ and $B$ respectively. This would violate the angle inequality $\angle E D C \neq 2 \cdot \angle A D B$ given in the question.
Next, let $E^{\prime}$ denote the second point of intersection of $P^{\prime} E$ with $\Gamma$, and let $E^{\prime \prime}$ denote the point on $\Gamma$ diametrically opposite $E^{\prime}$, so that $E^{\prime \prime} E$ is perpendicular to $P^{\prime} E$. The point $G$ lies on the perpendicular bisectors of the sides $P^{\prime} E$ and $E E^{\prime \prime}$ of the right-angled triangle $P^{\prime} E E^{\prime \prime}$; it follows that $G$ is the midpoint of $P^{\prime} E^{\prime \prime}$. Since $D$ is the midpoint of $E^{\prime} E^{\prime \prime}$, we have that $G D = \frac{1}{2} P^{\prime} E^{\prime}$. Since $P^{\prime} M = \frac{1}{2} P^{\prime} E$, we have $G D \cdot P^{\prime} M = \frac{1}{4} P^{\prime} E^{\prime} \cdot P^{\prime} E = \frac{1}{4} h$. The other equality $H D \cdot P^{\prime} N$ follows by exactly the same argument.
$\square$
From this claim, we see that the area condition is equivalent to the equality
$$\left(M A : P^{\prime} M\right) = \left(N B : P^{\prime} N\right)$$
of ratios of signed lengths, which is equivalent to direct similarity of $A P^{\prime} M$ and $B P^{\prime} N$, as desired.

Along the perpendicular bisector of $C E$, define the linear function
$$f(X) = \operatorname{area}(B C X) + \operatorname{area}(A X E) - \operatorname{area}(A B X) - \operatorname{area}(A B P),$$
where, from now on, we always use signed areas. Thus, we want to show that $C, P, E$ are collinear if and only if $f(D) = 0$.

Let $P^{\prime}$ be the reflection of $P$ across line $A B$. The point $P^{\prime}$ does not lie on the line $C E$. To see this, we let $A^{\prime \prime}$ and $B^{\prime \prime}$ be the points obtained from $A$ and $B$ by dilating with scale factor 2 about $P^{\prime}$, so that $P$ is the orthogonal projection of $P^{\prime}$ onto $A^{\prime \prime} B^{\prime \prime}$. Since $A$ lies on the perpendicular bisector of $P^{\prime} E$, the triangle $A^{\prime \prime} E P^{\prime}$ is right-angled at $E$ (and $B^{\prime \prime} C P^{\prime}$ similarly). If $P^{\prime}$ were to lie on $C E$, then the lines $A^{\prime \prime} E$ and $B^{\prime \prime} C$ would be perpendicular to $C E$ and $A^{\prime \prime}$ and $B^{\prime \prime}$ would lie on the opposite side of $C E$ to $D$. It follows that the line $A^{\prime \prime} B^{\prime \prime}$ does not meet triangle $C D E$, and hence point $P$ does not lie inside $C D E$. But then $P$ must lie inside $A B C E$, and it is clear that such a point cannot reflect to a point $P^{\prime}$ on $C E$.
We thus let $O$ be the centre of the circle $C E P^{\prime}$. The lines $A O$ and $B O$ are the perpendicular bisectors of $E P^{\prime}$ and $C P^{\prime}$, respectively, so
$$\begin{aligned} \operatorname{area}(B C O) + \operatorname{area}(A O E) & = \operatorname{area}\left(O P^{\prime} B\right) + \operatorname{area}\left(P^{\prime} O A\right) = \operatorname{area}\left(P^{\prime} B O A\right) \\ & = \operatorname{area}(A B O) + \operatorname{area}\left(B A P^{\prime}\right) = \operatorname{area}(A B O) + \operatorname{area}(A B P) \end{aligned}$$
and hence $f(O) = 0$.
Notice that if point $O$ coincides with $D$ then points $A, B$ lie in angle domain $C D E$ and $\angle E O C = 2 \cdot \angle A O B$, which is not allowed. So, $O$ and $D$ must be distinct. Since $f$ is linear and vanishes at $O$, it follows that $f(D) = 0$ if and only if $f$ is constant zero - we want to show this occurs if and only if $C, P, E$ are collinear.


In the one direction, suppose firstly that $C, P, E$ are not collinear, and let $T$ be the centre of the circle $C E P$. The same calculation as above provides
$$\operatorname{area}(B C T) + \operatorname{area}(A T E) = \operatorname{area}(P B T A) = \operatorname{area}(A B T) - \operatorname{area}(A B P)$$
so
$$f(T) = -2 \operatorname{area}(A B P) \neq 0 .$$
Hence, the linear function $f$ is nonconstant with its zero is at $O$, so that $f(D) \neq 0$.
In the other direction, suppose that the points $C, P, E$ are collinear. We will show that $f$ is constant zero by finding a second point (other than $O$ ) at which it vanishes.

Let $Q$ be the reflection of $P$ across the midpoint of $A B$, so $P A Q B$ is a parallelogram. It is easy to see that $Q$ is on the perpendicular bisector of $C E$; for instance if $A^{\prime}$ and $B^{\prime}$ are the points produced from $A$ and $B$ by dilating about $P$ with scale factor 2, then the projection of $Q$ to $C E$ is the midpoint of the projections of $A^{\prime}$ and $B^{\prime}$, which are $E$ and $C$ respectively. The triangles $B C Q$ and $A Q E$ are indirectly congruent, so
$$f(Q) = (\operatorname{area}(B C Q) + \operatorname{area}(A Q E)) - (\operatorname{area}(A B Q) - \operatorname{area}(B A P)) = 0 - 0 = 0 .$$
The points $O$ and $Q$ are distinct. To see this, consider the circle $\omega$ centred on $Q$ with $P^{\prime}$ on the circumference; since triangle $P P^{\prime} Q$ is right-angled at $P^{\prime}$, it follows that $P$ lies outside $\omega$. On the other hand, $P$ lies between $C$ and $E$ on the line $C P E$. It follows that $C$ and $E$ cannot both lie on $\omega$, so that $\omega$ is not the circle $C E P^{\prime}$ and $Q \neq O$.
Since $O$ and $Q$ are distinct zeroes of the linear function $f$, we have $f(D) = 0$ as desired.