IMO 2019 Shortlisted Problems

For the equivalence with the collinearity condition, let $F$ denote the foot of the perpendicular from $P^{\prime }$ to $AB$, so that $F$ is the midpoint of $P{P}^{\prime }$. We have that $P$ lies on $CE$ if and only if $F$ lies on $MN$, which occurs if and only if we have the equality $\angle AFM=\angle BFN$ of signed angles modulo $\pi$. By concyclicity of $A{P}^{\prime }FM$ and $BF{P}^{\prime }N$, 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 $\mathrm{area}(ABD)+\mathrm{area}(ABP)=\mathrm{area}\left(A{P}^{\prime }BD\right)=\mathrm{area}\left(A{P}^{\prime }D\right)+\mathrm{area}\left(BD{P}^{\prime }\right)$. Using the identity $\mathrm{area}(ADE)-\mathrm{area}\left(A{P}^{\prime }D\right)=\mathrm{area}(ADE)+\mathrm{area}\left(AD{P}^{\prime }\right)=2\mathrm{area}(ADM)$, and similarly for $B$, we find that the area condition is equivalent to the equality $$\mathrm{area}(DAM)=\mathrm{area}(DBN)$$ 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 $$MA\cdot GD=NB\cdot HD$$ (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 $DH$ 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 $GD$ and $HD$ 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 $$GD\cdot P^{\prime }M=HD\cdot P^{\prime }N=\frac{1}{4}h\ne 0.$$ Proof. Firstly, we have $h\ne 0$, since otherwise $P^{\prime }$ would lie on $\Gamma$, and hence the internal angle bisectors of $\angle ED{P}^{\prime }$ and $\angle P^{\prime }DC$ would pass through $A$ and $B$ respectively. This would violate the angle inequality $\angle EDC\ne 2\cdot \angle ADB$ 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 $GD=\frac{1}{2}{P}^{\prime }{E}^{\prime }$. Since $P^{\prime }M=\frac{1}{2}{P}^{\prime }E$, we have $GD\cdot P^{\prime }M=\frac{1}{4}{P}^{\prime }{E}^{\prime }\cdot P^{\prime }E=\frac{1}{4}h$. The other equality $HD\cdot P^{\prime }N$ follows by exactly the same argument. $\square$ From this claim, we see that the area condition is equivalent to the equality $$\left(MA:P^{\prime }M\right)=\left(NB:P^{\prime }N\right)$$ of ratios of signed lengths, which is equivalent to direct similarity of $A{P}^{\prime }M$ and $B{P}^{\prime }N$, as desired.

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Alternative solution.

Along the perpendicular bisector of $CE$, define the linear function $$f(X)=\mathrm{area}(BCX)+\mathrm{area}(AXE)-\mathrm{area}(ABX)-\mathrm{area}(ABP),$$ 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 $AB$. The point $P^{\prime }$ does not lie on the line $CE$. 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 $CE$, then the lines $A^{\prime \prime }E$ and $B^{\prime \prime }C$ would be perpendicular to $CE$ and $A^{\prime \prime }$ and $B^{\prime \prime }$ would lie on the opposite side of $CE$ to $D$. It follows that the line $A^{\prime \prime }{B}^{\prime \prime }$ does not meet triangle $CDE$, and hence point $P$ does not lie inside $CDE$. But then $P$ must lie inside $ABCE$, and it is clear that such a point cannot reflect to a point $P^{\prime }$ on $CE$. We thus let $O$ be the centre of the circle $CE{P}^{\prime }$. The lines $AO$ and $BO$ are the perpendicular bisectors of $E{P}^{\prime }$ and $C{P}^{\prime }$, respectively, so $$\begin{array}{rl}\mathrm{area}(BCO)+\mathrm{area}(AOE)& =\mathrm{area}\left(O{P}^{\prime }B\right)+\mathrm{area}\left(P^{\prime }OA\right)=\mathrm{area}\left(P^{\prime }BOA\right)\\ & =\mathrm{area}(ABO)+\mathrm{area}\left(BA{P}^{\prime }\right)=\mathrm{area}(ABO)+\mathrm{area}(ABP)\end{array}$$ and hence $f(O)=0$. Notice that if point $O$ coincides with $D$ then points $A,B$ lie in angle domain $CDE$ and $\angle EOC=2\cdot \angle AOB$, 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 $CEP$. The same calculation as above provides $$\mathrm{area}(BCT)+\mathrm{area}(ATE)=\mathrm{area}(PBTA)=\mathrm{area}(ABT)-\mathrm{area}(ABP)$$ so $$f(T)=-2\mathrm{area}(ABP)\ne 0.$$ Hence, the linear function $f$ is nonconstant with its zero is at $O$, so that $f(D)\ne 0$. 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 $AB$, so $PAQB$ is a parallelogram. It is easy to see that $Q$ is on the perpendicular bisector of $CE$; 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 $CE$ is the midpoint of the projections of $A^{\prime }$ and $B^{\prime }$, which are $E$ and $C$ respectively. The triangles $BCQ$ and $AQE$ are indirectly congruent, so $$f(Q)=(\mathrm{area}(BCQ)+\mathrm{area}(AQE))-(\mathrm{area}(ABQ)-\mathrm{area}(BAP))=0-0=0.$$ 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 $CPE$. It follows that $C$ and $E$ cannot both lie on $\omega$, so that $\omega$ is not the circle $CE{P}^{\prime }$ and $Q\ne O$. Since $O$ and $Q$ are distinct zeroes of the linear function $f$, we have $f(D)=0$ as desired.