, 2023

Figure 1: Diagram to solution 1

Let $M$ be the midpoint of $K L$. We will prove that $M$ is the orthocentre of $A B C$. Since $D K$ and $D L$ are tangent to the same circle, $|D K|=|D L|$ and hence $D M \perp K L$. The theorem of Thales in circle $A B C$ also gives $D B \perp B A$ and $D C \perp C A$. The right angles then give that quadrilaterals $B D M K$ and $D M L C$ are cyclic.
If $\angle B A C=\alpha$, then clearly $\angle D K M=\angle M L D=\alpha$ by angle in the alternate segment of circle $A K L$, and so $\angle M D K=\angle L D M=\frac{\pi}{2}-\alpha$, which thanks to cyclic quadrilaterals gives $\angle M B K=\angle L C M=\frac{\pi}{2}-\alpha$. From this, we have $B M \perp A C$ and $C M \perp A B$, and so $M$ indeed is the orthocentre of $A B C$.

## Preliminaries
Let $A B C$ be a triangle with circumcircle $\Gamma$. Let $X$ be a point in the plane. The Simson line (Wallace-Simson line) is defined via the following theorem. Drop perpendiculars from $X$ to each of the three side lines of $A B C$. The feet of these perpendiculars are collinear (on the Simson line of $X$) if and only if $X$ lies on $\Gamma$. The Simson line of $X$ in the circumcircle bisects the line segment $X H$ where $H$ is the orthocentre of triangle $A B C$. See Figure 2

Figure 2: The Wallace-Simson configuration

When $X$ is on $\Gamma$, we can enlarge from $X$ with scale factor 2 (a homothety) to take the Simson line to the doubled Simson line which passes through the orthocentre $H$ and contains the reflections of $X$ in each of the three sides of $A B C$.

## Solution of the problem

Figure 3: Three circles do the work

Let $\Gamma$ be the circle $A B C$, $\Sigma$ be the circle $A K L$ with centre $O$, and $\Omega$ be the circle on diameter $O D$ so $K$ and $L$ are on this circle by converse of Thales. Let $\Omega$ and $\Gamma$ meet at $D$ and $F$. By Thales in both circles, $\angle A F D$ and $\angle O F D$ are both right angles so $A O F$ is a line. Let $A F$ meet $\Sigma$ again at $T$ so $A T$ (containing $O$) is a diameter of this circle and by Thales, $T L \perp A C$.

Let $G$ (on $\Sigma$) be the reflection of $K$ in $A F$. Now $A T$ is the internal angle bisector of $\angle G A K$ so, by an upmarket use of angles in the same segment (of $\Sigma$), $T L$ is the internal angle bisector of $\angle G L K$. Thus the line $G L$ is the reflection of the line $K L$ in $T L$, and so also the reflection of $K L$ in the line $A C$ (internal and external angle bisectors).

Our next project is to show that $L G F$ are collinear. Well $\angle F L K=\angle F O K$ (angles in the same segment of $\Omega$) and $\angle G L K=\angle G A K$ (angles in the same segment of $\Sigma$) $=2 \angle O A K$ ($A K G$ is isosceles with apex $A)=\angle T O K$ (since $O A K$ is isosceles with apex $O$, and this is an external angle at $O$). The point $T$ lies in the interior of the line segment $F O$ so $\angle T O K=\angle F O K$. Therefore $\angle F L K=\angle G L K$ so $L G F$ is a line.

Now from the second paragraph, $F$ is on the reflection of $K L$ in $A C$. By symmetry, $F$ is also on the reflection of $K L$ in $A B$. Therefore the reflections of $F$ in $A B$ and $A C$ are both on $K L$ which must therefore be the doubled Wallace-Simson line of $F$. Therefore the orthocentre of $A B C$ lies on $K L$.

Let $H$ be the orthocentre of triangle $A B C$ and $\Sigma$ the circumcircle of $A K L$ with centre $O$. Let $\Omega$ be the circle with diameter $O D$, which contains $K$ and $L$ by Thales, and let $\Gamma$ be the circumcircle of $A B C$ containing $D$. Denote the second intersection of $\Omega$ and $\Gamma$ by $F$. Since $O D$ and $A D$ are diameters of $\Omega$ and $\Gamma$ we have $\angle O F D=\frac{\pi}{2}=\angle A F D$, so the points $A, O, F$ are collinear. Let $M$ and $N$ be the second intersections of $C H$ and $B H$ with $\Gamma$, respectively. It is well-known that $M$ and $N$ are the reflections of $H$ in $A B$ and $A C$, respectively (because $\angle N C A=\angle N B A=\angle A C M=\angle A B M$). By collinearity of $A, O, F$ and the angles in $\Gamma$ we have
$$\angle N F O=\angle N F A=\angle N B A=\frac{\pi}{2}-\angle B A C=\frac{\pi}{2}-\angle K A L$$
Since $D L$ is tangent to $\Sigma$ we obtain
$$\angle N F O=\frac{\pi}{2}-\angle K L D=\angle L D O,$$
where the last equality follows from the fact that $O D$ is bisector of $\angle L D K$ since $L D$ and $K D$ are tangent to $\Sigma$. Furthermore, $\angle L D O=\angle L F O$ since these are angles in $\Omega$. Hence, $\angle N F O=\angle L F O$, which implies that points $N, L, F$ are collinear. Similarly points $M, K, F$ are collinear. Since $N$ and $M$ are reflections of $H$ in $A C$ and $A B$ we have
$$\angle L H N=\angle H N L=\angle B N F=\angle B M F=\angle B M K=\angle K H B.$$
Hence,
$$\angle L H K=\angle L H N+\angle N H K=\angle K H B+\angle N H K=\pi$$
and the points $L, H, K$ are collinear.

As in Solution 3 let $M$ and $N$ be the reflections of the orthocentre in $A B$ and $A C$. Let $\angle B A C=\alpha$. Then $\angle N D M=\pi-\angle M A N=\pi-2 \alpha$.
Let $M K$ and $N L$ intersect at $F$. See Figure 3.

Claim. $\angle N F M=\pi-2 \alpha$, so $F$ lies on the circumcircle.

Proof. Since $K D$ and $L D$ are tangents to circle $A K L$, we have $|D K|=|D L|$ and $\angle D K L=\angle K L D=\alpha$, so $\angle L D K=\pi-2 \alpha$.
By definition of $M, N$ and $D, \angle M N D=\angle A N D-\angle A N M=\frac{\pi}{2}-\left(\frac{\pi}{2}-\alpha\right)=\alpha$ and analogously $\angle D M N=\alpha$. Hence $|D M|=|D N|$.
From $\angle N D M=\angle L D K=\pi-2 \alpha$ it follows that $\angle L D N=\angle K D M$. Since $|D K|=|D L|$ and $|D M|=|D N|$, triangles $M D K$ and $N D L$ are related by a rotation about $D$ through angle $\pi-2 \alpha$, and hence the angle between $M K$ and $N L$ is $\pi-2 \alpha$, which proved the claim.

We now finish as in Solution 3:
$$\begin{gathered} \angle M H K=\angle K M H=\angle F M C=\angle F A C \\ \angle L H N=\angle H N L=\angle B N F=\angle B A F \end{gathered}$$
As $\angle B A F+\angle F A C=\alpha$, we have $\angle L H K=\alpha+\angle N H M=\alpha+\pi-\alpha=\pi$, so $H$ lies on $K L$.

Since $A D$ is a diameter, it is well known that $D B H C$ is a parallelogram (indeed, both $B D$ and $C H$ are perpendicular to $A B$, hence parallel, and similarly for $D C \parallel B H$). Let $B'$, $C'$ be the reflections of $D$ in lines $A K B$ and $A L C$, respectively; since $A B D$ and $A C D$ are right angles, these are also the factor-2 homotheties of $B$ and $C$ with respect to $D$, hence $H$ is the midpoint of $B' C'$. We will prove that $B' K C' L$ is a parallelogram: it will then follow that the midpoint of $B' C'$, which is $H$, is also the midpoint of $K L$, and in particular is on the line, as we wanted to show.

We will prove $B' K C' L$ is a parallelogram by showing that $B' K$ and $C' L$ are the same length and direction. Indeed, for lengths we have $K B'=K D=L D=L C'$, where the first and last equalities arise from the reflections defining $B'$ and $C'$, and the middle one is equality of tangents. For directions, let $\alpha, \beta, \gamma$ denote the angles of triangle $A K L$. Immediate angle chasing in the circle $A K L$, and the properties of the reflections, yield
$$\begin{aligned} \angle C' L C &=\angle C L D=\angle A K L=\beta \\ \angle B K B' &=\angle D K B=\angle K L A=\gamma \\ & \angle L D K=2 \alpha-\pi \end{aligned}$$
and therefore in directed angles $(\bmod 2 \pi)$ we have
$\angle\left(C' L, B' K\right)=\angle C' L C+\angle C L D+\angle L D K+\angle D K B+\angle B K B'=2 \alpha+2 \beta+2 \gamma-\pi=\pi$
and hence $C' L$ and $B' K$ are parallel and in opposite directions, i.e. $C' L$ and $K B'$ are in the same direction, as claimed.

There are a number of "phantom point" arguments which define $K'$ and $L'$ in terms of angles and then deduce that these points are actually $K$ and $L$.

Note: In these solutions it is necessary to show that $K$ and $L$ are uniquely determined by the conditions of the problem. One example of doing this is the following:
To prove uniqueness of $K$ and $L$, let us consider that there exist two other points $K'$ and $L'$ that satisfy the same properties ($K'$ on $A B$ and $L'$ on $A C$ such that $D K'$ and $D L'$ are tangent to the circle $A K' L'$).
Then, we have that $D K=D L$ and $D K'=D L'$. We also have that $\angle K D L=\angle K' D L'=\pi-2 \angle A$. Hence, we deduce $\angle K D K'=\angle K D L-\angle K' D L=\angle K' D L'-\angle K' D L=\angle L D L'$ Thus we have that $\triangle K D K' \equiv \triangle L D L'$, so we deduce $\angle D K A=\angle D K K'=\angle D L L'=\pi-\angle A L D$. This implies that $A K D L$ is concyclic, which is clearly a contradiction since $\angle K A L+\angle K D L=\pi-\angle B A C$.

We will use the usual complex number notation, where we will use a capital letter (like $Z$) to denote the point associated to a complex number (like $z$). Consider $\triangle A K L$ on the unit circle. So, we have $a \cdot \bar{a}=k \cdot \bar{k}=l \cdot \bar{l}=1$. As point $D$ is the intersection of the tangents to the unit circle at $K$ and $L$, we have that
$$d=\frac{2 k l}{k+l} \text{ and } \bar{d}=\frac{2}{k+l}$$
Defining $B$ as the foot of the perpendicular from $D$ on the line $A K$, and $C$ as the foot of the perpendicular from $D$ on the line $A L$, we have the formulas:
$$b=\frac{1}{2}\left(d+\frac{(a-k) \bar{d}+\bar{a} k-a \bar{k}}{\bar{a}-\bar{k}}\right)$$
$$c=\frac{1}{2}\left(d+\frac{(a-l) \bar{d}+\bar{a} l-a \bar{l}}{\bar{a}-\bar{l}}\right)$$
Simplyfing these formulas, we get:
$$\begin{gathered} b=\frac{1}{2}\left(d+\frac{(a-k) \frac{2}{k+l}+\frac{k}{a}-\frac{a}{k}}{\frac{1}{a}-\frac{1}{k}}\right)=\frac{1}{2}\left(d+\frac{\frac{2(a-k)}{k+l}+\frac{k^2-a^2}{a k}}{\frac{k-a}{a k}}\right) \\ b=\frac{1}{2}\left(\frac{2 k l}{k+l}-\frac{2 a k}{k+l}+(a+k)\right)=\frac{k(l-a)}{k+l}+\frac{1}{2}(k+a) \\ c=\frac{1}{2}\left(d+\frac{(a-l) \frac{2}{k+l}+\frac{l}{a}-\frac{a}{l}}{\frac{1}{a}-\frac{1}{l}}\right)=\frac{1}{2}\left(d+\frac{\frac{2(a-l)}{k+l}+\frac{l^2-a^2}{a l}}{\frac{l-a}{a l}}\right) \\ c=\frac{1}{2}\left(\frac{2 k l}{k+l}-\frac{2 a l}{k+l}+(a+l)\right)=\frac{l(k-a)}{k+l}+\frac{1}{2}(l+a) \end{gathered}$$
Let $O$ be the circumcenter of triangle $\triangle A B C$. As $A D$ is the diameter of this circle, we have that:
$$o=\frac{a+d}{2}$$
Defining $H$ as the orthocentre of the $\triangle A B C$, we get that:
$$\begin{gathered} h=a+b+c-2 \cdot o=a+\left(\frac{k(l-a)}{k+l}+\frac{1}{2}(k+a)\right)+\left(\frac{l(k-a)}{k+l}+\frac{1}{2}(l+a)\right)-(a+d) \\ h=a+\frac{2 k l}{k+l}-\frac{a(k+l)}{k+l}+\frac{1}{2} k+\frac{1}{2} l+a-\left(a+\frac{2 k l}{k+l}\right) \\ h=\frac{1}{2}(k+l) \end{gathered}$$
Hence, we conclude that $H$ is the midpoint of $K L$, so $H, K, L$ are collinear.

Let us employ the barycentric coordinates. Set $A(1,0,0), K(0,1,0), L(0,0,1)$.
The tangent at $K$ of $(A K L)$ is $a^{2} z+c^{2} x=0$, and the tangent at $L$ of $(A K L)$ is $a^{2} y+b^{2} x=0$. Their intersection is
$$D\left(-a^{2}: b^{2}: c^{2}\right)$$
Since $B \in A K$, we can let $B(1-t, t, 0)$. Solving for $\overrightarrow{A B} \cdot \overrightarrow{B D}=0$ gives
$$t=\frac{3 b^{2}+c^{2}-a^{2}}{2\left(b^{2}+c^{2}-a^{2}\right)} \Longrightarrow B=\left(\frac{-a^{2}-b^{2}+c^{2}}{2\left(b^{2}+c^{2}-a^{2}\right)}, \frac{-a^{2}+3 b^{2}+c^{2}}{2\left(b^{2}+c^{2}-a^{2}\right)}, 0\right)$$
Likewise, $C$ has the coordinate
$$C=\left(\frac{-a^{2}+b^{2}-c^{2}}{2\left(b^{2}+c^{2}-a^{2}\right)}, 0, \frac{-a^{2}+b^{2}+3 c^{2}}{2\left(b^{2}+c^{2}-a^{2}\right)}\right)$$
The altitude from $B$ for triangle $A B C$ is
$$-b^{2}\left(x-z-\frac{-a^{2}-b^{2}+c^{2}}{2\left(b^{2}+c^{2}-a^{2}\right)}\right)+\left(c^{2}-a^{2}\right)\left(y-\frac{-a^{2}+3 b^{2}+c^{2}}{2\left(b^{2}+c^{2}-a^{2}\right)}\right)=0$$
Also the altitude from $C$ for triangle $A B C$ is
$$-c^{2}\left(x-y-\frac{-a^{2}+b^{2}-c^{2}}{2\left(b^{2}+c^{2}-a^{2}\right)}\right)+\left(a^{2}-b^{2}\right)\left(z-\frac{-a^{2}+b^{2}+3 c^{2}}{2\left(b^{2}+c^{2}-a^{2}\right)}\right)=0$$
The intersection of these two altitudes, which is the orthocenter of triangle $A B C$, has the barycentric coordinate
$$H=(0,1/2,1/2)$$
which is the midpoint of the segment $K L$.