Subfield: Geometry


Question: Let $A B C D E$ be a convex pentagon with $C D=D E$ and $\angle E D C \neq 2 \cdot \angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P=A E$ and $B P=B C$. Prove that $P$ lies on the diagonal $C E$ if and only if area $(B C D)+\operatorname{area}(A D E)=$ $\operatorname{area}(A B D)+\operatorname{area}(A B P)$.

Solution:
Let $P^{\prime}$ be the reflection of $P$ across line $A B$, and let $M$ and $N$ be the midpoints of $P^{\prime} E$ and $P^{\prime} C$ respectively. Convexity ensures that $P^{\prime}$ is distinct from both $E$ and $C$, and hence from both $M$ and $N$. We claim that both the area condition and the collinearity condition in the problem are equivalent to the condition that the (possibly degenerate) right-angled triangles $A P^{\prime} M$ and $B P^{\prime} N$ are directly similar (equivalently, $A P^{\prime} E$ and $B P^{\prime} C$ are directly similar).

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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.

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For the other equivalence with the area condition, we have the equality of signed areas 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 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

$$
M A \cdot G D=N B \cdot H D \text {. }
$$

(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 $B$.)



<img_3246>

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.

<img_3232>

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 $f(D)=0$.

<img_3489>

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.
<img_3610>

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

$$
\operatorname{area}(B C T)+\operatorname{area}(A T E)=\operatorname{area}(P B T A)=\operatorname{area}(A B T)-\operatorname{area}(A B P)
$$

$$
f(T)=-2 \operatorname{area}(A B P) \neq 0 .
$$



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.

<img_3865>

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.