Subfield: Geometry


Question: Consider the triangle $A B C$ with $\angle B C A>90^{\circ}$. The circumcircle $\Gamma$ of $A B C$ has radius $R$. There is a point $P$ in the interior of the line segment $A B$ such that $P B=P C$ and the length of $P A$ is $R$. The perpendicular bisector of $P B$ intersects $\Gamma$ at the points $D$ and $E$.

Prove that $P$ is the incentre of triangle $C D E$.

Solution:
The angle bisector of $\angle E C D$ intersects the circumcircle of $C D E$ (which is $\Gamma$ ) at the midpoint $M$ of arc $D B E$. It is well-known that the incentre is the intersection of the angle bisector segment $C M$ and the circle with centre at $M$ and passing through $D, E$. We will verify this property for $P$.

<img_4013>

By the conditions we have $A P=O A=O B=O C=O D=O E=R$. Both lines $O M$ and $A P B$ are perpendicular to $E D$, therefore $A P \| O M$; in the quadrilateral $A O M P$ we have $A P=O A=A M=R$ and $A P \| O M$, so $A O M P$ is a rhombus and its fourth side is $P M=R$. In the convex quadrilateral $O M B P$ we have $O M \| P B$, so $O M B P$ is a symmetric trapezoid; the perpendicular bisector of its bases $A O$ and $P B$ coincide. From this symmetry we obtain $M D=O D=R$ and $M E=O E=R$. (Note that the triangles $O E M$ ad $O M D$ are equilateral.) We already have $M P=M D=M E=R$, so $P$ indeed lies on the circle with center $M$ and passing through $D, E$. (Notice that this circle is the reflection of $\Gamma$ about $D E$.)

From $P B=P C$ and $O B=O C$ we know that $B$ and $C$ are symmetrical about $O P$; from the rhombus $A O M P$ we find that $A$ and $M$ are also symmetrical about $O P$. By reflecting the collinear points $B, P, A$ (with $P$ lying in the middle) we get that $C, P, M$ are collinear (and $P$ is in the middle). Hence, $P$ lies on the line segments $C M$.
Let $X$ be the second intersection of $C P$ with $\Gamma$. Using the power of the point $P$ in the circle $\Gamma$ and the fact that $P B=P C$, we find that $P X=P A=R$. The quadrilateral $A O X P$ has four sides of equal length, so it is a rhombus and in particular $O X$ is parallel to $A P$. This proves that $O X B P$ is a trapezoid, and because the diagonals $P X$ and $O B$ have equal length, this is even an isosceles trapezoid. Because of that, $D E$ is not only the perpendicular bisector of $P B$, but also of $O X$.

<img_3704>

In particular we have $X D=X P=X O=X E=R$, which proves that $X$ is the middle of the arc $D E$ and $P$ belongs to the circle with center $X$ going through $D$ and $E$. These properties, together with the fact that $C, P, X$ are collinear, determine uniquely the incenter of $C D E$.
Let $Y$ be the circumcenter of triangle $B P C$. Then from $Y B=Y P$ it follows that $Y$ lies on $D E$ (we assume $D$ lies in between $Y$ and $E$ ), and from $Y B=Y C$ it follows that $Y$ lies on $O P$, where $O$ is the center of $\Gamma$.

From $\angle A O C=2 \angle A B C=\angle A P C$ (because $\angle P B C=\angle P C B)$ we deduce that $A O P C$ is a cyclic quadrilateral, and from $A P=R$ it follows that $A O P C$ is an isosceles trapezoid. We now find that $\angle Y C P=\angle Y P C=180^{\circ}-\angle O P C=180^{\circ}-\angle A C P$, so $Y$ lies on $A C$.

<img_3430>

Power of a point gives $Y O \cdot Y P=Y C \cdot Y A=Y D \cdot Y E$, so $D, P, O$ and $E$ are concyclic. It follows that $2 \angle D A E=\angle D O E=\angle D P E=\angle D B E=180^{\circ}-\angle D A E$, so $\angle D A E=60^{\circ}$. We can now finish the proof by angle chasing.

From $A B \perp D E$ we have $\angle A O D+\angle B O E=180^{\circ}$ and from $\angle D O E=2 \angle D A E=120^{\circ}$ it follows that $\angle B O D+\angle B O E=120^{\circ}$. It follows that $\angle A O D-\angle B O D=180^{\circ}-120^{\circ}=60^{\circ}$. Let $\angle O A B=\angle O B A=2 \beta$; then $\angle A O D+\angle B O D=\angle A O B=180^{\circ}-4 \beta$. Together with $\angle A O D-\angle B O D=60^{\circ}$, this yields $\angle A O D=120^{\circ}-2 \beta$ and $\angle B O D=60^{\circ}-2 \beta$. We now find $\angle A E D=\frac{1}{2} \angle A O D=60^{\circ}-\beta$, which together with $\angle D A E=60^{\circ}$ yields $\angle A D E=60^{\circ}+\beta$. From the isosceles trapezoid $A O P C$ we have $\angle C D A=\angle C B A=\frac{1}{2} \angle C P A=\frac{1}{2} \angle P A O=\beta$, so $\angle C D E=\angle C D A+\angle A D E=\beta+60^{\circ}+\beta=60^{\circ}+2 \beta$.

From $\angle B O D=60^{\circ}-2 \beta$ we deduce that $\angle B E D=30^{\circ}-\beta$; together with $\angle D B E=120^{\circ}$ this yields $\angle E D B=30^{\circ}+\beta$. We now see that $\angle P D E=\angle B D E=30^{\circ}+\beta=\frac{1}{2} \angle C D E$, so $P$ is on the angle bisector of $\angle C D E$. Similarly, $P$ lies on the angle bisector of $\angle C E D$, so $P$ is the incenter of $C D E$.
We draw the lines $D P$ and $E P$ and let $D^{\prime}$ resp. $E^{\prime}$ be the second intersection point with $\Gamma$.

<img_3206>

The triangles $A P D^{\prime}$ and $D P B$ are similar, and the triangles $A P E^{\prime}$ and $E P B$ are also similar, hence they are all isosceles and it follows that $E^{\prime}, O, P, D^{\prime}$ lie on a circle with center $A$. In particular $A O D^{\prime}$ and $A O E^{\prime}$ are equilateral triangles. Angle chasing gives

$$
\begin{gathered}
\angle C D P=\angle C D D^{\prime}=\frac{1}{2} \angle C O D^{\prime}=\frac{1}{2}\left(60^{\circ}+\angle C O A\right) \\
\angle E D P=\angle E D D^{\prime}=\angle E E^{\prime} D^{\prime}=\angle P E^{\prime} D^{\prime}=\frac{1}{2} \angle P A D^{\prime}=\frac{1}{2}\left(60^{\circ}+\angle P A O\right)
\end{gathered}
$$

Similarly we prove $\angle C E P=\frac{1}{2}\left(60^{\circ}-\angle C O A\right)$ and $\angle D E P=\frac{1}{2}\left(60^{\circ}-\angle P A O\right)$ so if we can prove that $\angle C O A=\angle P A O$, we will have proven that $P$ belongs to the angle bisector of $\angle C E D$ and to the angle bisector of $\angle C D E$, which is enough to prove that $P$ is the incenter of the triangle $C D E$.

Let $\beta=\angle A B C=\angle P C B$. We have $\angle A P C=2 \beta$ and $\angle A O C=2 \beta$, so $A O P C$ is an inscribed quadrilateral. Moreover, since the diagonals $A P$ and $C O$ have equal length, this is actually an isosceles trapezoid, and hence $\angle P A O=\angle C P A=\angle C O A$ which concludes the proof.
Without loss of generality we assume that $D$ and $C$ are in the same half-plane regarding line $A B$.

Since $P C=B P$ and $A B C D$ is inscribed quadrilateral we have $\angle P C B=\angle C B P=\angle C E A=\alpha$. As in the other solutions, $A O P C$ is an isosceles trapezoid and $2 \alpha=\angle C P A=\angle P C O$

<img_3821>

Let $K$ be intersection of $E O$ and $\Gamma$. Then $\angle K D E=90^{\circ}, A B \| D K$ and $K D B A$ is isosceles trapezoid. We obtain $D P=B D=A K$, which implies that $D P A K$ is a parallelogram and hence $D K=P A=R=O D=O K$. We see that $D O K$ is an equilateral triangle. Then $\angle E C D=\angle E K D=60^{\circ}$.

Further we prove that $P C$ bisects $\angle E C D$ using $\angle D C B=\angle D K B=\angle K D A$ (from isosceles trapeziod $D K A B$ ) and that $\angle K E C=\angle O E C=\angle O C E$ (from isosceles triangle $O C E)$ :

$$
\angle D C P=\angle D C B+\angle B C P=\angle K D A+\alpha=\angle K E C+\angle C E A+\alpha=\angle O C E+2 \alpha=\angle P C E
$$

Further by $\angle D C P=\angle P C E=\frac{1}{2} \angle E C D=30^{\circ}$ distances between $P$ and sides $\triangle C D E$ are $\frac{1}{2} P C=\frac{1}{2} P B$ (as ratio between cathetus and hypotenuse in right triangle with angles $60^{\circ}$ and $30^{\circ}$ ). So, we have found incentre.
Assume $A B$ is parallel to the horizontal axis, and that $\Gamma$ is the unit circle. Write $f(\theta)$ for the point $(\cos (\theta), \sin (\theta))$ on $\Gamma$. As in Solution $\mathrm{C}$, assume that $\angle A O B=180^{\circ}-4 \beta$; then we can take $B=f(2 \beta)$ and $A=f\left(180^{\circ}-2 \beta\right)$. We observe that $A O P C$ is an isosceles trapezoid, which we use to deduce that $\angle A B C=\frac{1}{2} \angle A P C=\frac{1}{2} \angle O A B=\beta$. We now know that $C=f\left(180^{\circ}-4 \beta\right)$.

The point $P$ lies on $A B$ with $A P=R=1$, so $P=\left(\cos \left(180^{\circ}-2 \beta\right)+1, \sin (2 \beta)\right)=(1-$ $\cos (2 \beta), \sin (2 \beta))$. The midpoint of $B P$ therefore has coordinates $\left(\frac{1}{2}, \sin (2 \beta)\right)$, so $D$ and $E$ have $x$-coordinate $\frac{1}{2}$. Without loss of generality, we take $D=f\left(60^{\circ}\right)$ and $E=f\left(-60^{\circ}\right)$.

We have now obtained coordinates for all points in the problem, with one free parameter $(\beta)$. To show that $P$ is the incenter of $C D E$, we will show that $P$ lies on the bisector of $\angle C D E$; analogously, one can show that $P$ lies on the bisector of angle $C E D$. The bisector of angle $C D E$ passes through the midpoint $M$ of the arc $C E$ not containing $D E$; because $C=f\left(180^{\circ}-4 \beta\right)$ and $E=f\left(-60^{\circ}\right)$, we have $M=f\left(240^{\circ}-2 \beta\right)$.

It remains to show that $P=(1-\cos (2 \beta), \sin (2 \beta))$ lies on the line connecting the points $D=$ $\left(\cos \left(60^{\circ}\right), \sin \left(60^{\circ}\right)\right)$ and $M=\left(\cos \left(240^{\circ}-2 \beta\right), \sin \left(240^{\circ}-2 \beta\right)\right)=\left(-\cos \left(60^{\circ}-2 \beta\right),-\sin \left(60^{\circ}-2 \beta\right)\right)$. The equation for the line $D M$ is

$\left(Y+\sin \left(60^{\circ}-2 \beta\right)\right)\left(\cos \left(60^{\circ}\right)+\cos \left(60^{\circ}-2 \beta\right)\right)=\left(\sin \left(60^{\circ}\right)+\sin \left(60^{\circ}-2 \beta\right)\right)\left(X+\cos \left(60^{\circ}-2 \beta\right)\right)$,

which, using the fact that $\cos \left(60^{\circ}\right)+\cos \left(60^{\circ}-2 \beta\right)=2 \cos (\beta) \cos \left(60^{\circ}-\beta\right)$ and $\sin \left(60^{\circ}\right)+\sin \left(60^{\circ}-\right.$ $2 \beta)=2 \cos (\beta) \sin \left(60^{\circ}-\beta\right)$, simplifies to

$$
\left(Y+\sin \left(60^{\circ}-2 \beta\right)\right) \cos \left(60^{\circ}-\beta\right)=\left(X+\cos \left(60^{\circ}-2 \beta\right)\right) \sin \left(60^{\circ}-\beta\right) \text {. }
$$

Because $\cos \left(60^{\circ}-2 \beta\right) \sin \left(60^{\circ}-\beta\right)-\sin \left(60^{\circ}-2 \beta\right) \cos \left(60^{\circ}-\beta\right)=\sin (\beta)$, this equation further simplifies to

$$
Y \cos \left(60^{\circ}-\beta\right)-X \sin \left(60^{\circ}-\beta\right)=\sin (\beta) .
$$



Plugging in the coordinates of $P$, i.e., $X=1-\cos (2 \beta)$ and $Y=\sin (2 \beta)$, shows that $P$ is on this line: for this choice of $X$ and $Y$, the left hand side equals $\sin \left(60^{\circ}+\beta\right)-\sin \left(60^{\circ}-\beta\right)=$ $2 \cos \left(60^{\circ}\right) \sin (\beta)$, which is indeed equal to $\sin (\beta)$. So $P$ lies on the bisector $D M$ of $\angle C D E$, as desired.
Let $\Gamma$ be the complex unit circle and let $A B$ be parallel with the real line and $0<\varphi=\arg b<\frac{\pi}{2}$. Then

$$
|b|=1, \quad a=-\bar{b}, \quad p=a+1=1-\bar{b}
$$

From $\operatorname{Re} d=\operatorname{Re} e=\operatorname{Re} \frac{p+b}{2}=\frac{1}{2}$ we get that $d=\frac{1}{2}+\frac{\sqrt{3}}{2} i$ and $e=\frac{1}{2}-\frac{\sqrt{3}}{2} i$ are conjugate 6th roots of unity; $d^{3}=e^{3}=-1, d+e=1, d^{2}=-e, e^{2}=-d$ etc.

Point $C$ is the reflection of $B$ in line $O P$. From $\arg p=\arg (1-\bar{b})=\frac{1}{2}(\pi-\varphi)$, we can get $\arg c=2 \arg p-\arg b=\pi-2 \varphi$, so $c=-\bar{b}^{2}$.

Now we can verify that $E P$ bisects $\angle C E D$. This happens if and only if $(p-e)^{2}(\bar{c}-\bar{e})(\bar{d}-\bar{e})$ is real. Since $\bar{d}-\bar{e}=-\sqrt{3} i$, this is equivalent with $\operatorname{Re}\left[(p-e)^{2}(\bar{c}-\bar{e})\right]=0$. Here

$$
\begin{aligned}
(p-e)^{2}(\bar{c}-\bar{e}) & =(1-\bar{b}-e)^{2}\left(-b^{2}-d\right)=(d-\bar{b})^{2}\left(-b^{2}-d\right) \\
& =-|b|^{4}+2 d|b|^{2} b-d^{2} b^{2}-d \bar{b}^{2}+2 d^{2} \bar{b}-d^{3} \\
& =-1-2 d b+\bar{d} b^{2}-d \bar{b}^{2}-2 \overline{d b}+1 \\
& =-2(d b-\overline{d b})+\left(\bar{d} b^{2}-d \bar{b}^{2}\right)
\end{aligned}
$$

whose real part is zero. It can be proved similarly that $D P$ bisects $\angle E D C$.