Subfield: Geometry


Question: Let $A B C$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $A C$ and $A B$, respectively, and let $M$ be the midpoint of $E F$. Let the perpendicular bisector of $E F$ intersect the line $B C$ at $K$, and let the perpendicular bisector of $M K$ intersect the lines $A C$ and $A B$ at $S$ and $T$, respectively. We call the pair $(E, F)$ interesting, if the quadrilateral $K S A T$ is cyclic.

Suppose that the pairs $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ are interesting. Prove that

$$
\frac{E_{1} E_{2}}{A B}=\frac{F_{1} F_{2}}{A C}
$$

Solution:
For any interesting pair $(E, F)$, we will say that the corresponding triangle $E F K$ is also interesting.

Let $E F K$ be an interesting triangle. Firstly, we prove that $\angle K E F=\angle K F E=\angle A$, which also means that the circumcircle $\omega_{1}$ of the triangle $A E F$ is tangent to the lines $K E$ and $K F$.

Denote by $\omega$ the circle passing through the points $K, S, A$, and $T$. Let the line $A M$ intersect the line $S T$ and the circle $\omega$ (for the second time) at $N$ and $L$, respectively (see Figure 1).

Since $E F \| T S$ and $M$ is the midpoint of $E F, N$ is the midpoint of $S T$. Moreover, since $K$ and $M$ are symmetric to each other with respect to the line $S T$, we have $\angle K N S=\angle M N S=$ $\angle L N T$. Thus the points $K$ and $L$ are symmetric to each other with respect to the perpendicular bisector of $S T$. Therefore $K L \| S T$.

Let $G$ be the point symmetric to $K$ with respect to $N$. Then $G$ lies on the line $E F$, and we may assume that it lies on the ray $M F$. One has

$$
\angle K G E=\angle K N S=\angle S N M=\angle K L A=180^{\circ}-\angle K S A
$$

(if $K=L$, then the angle $K L A$ is understood to be the angle between $A L$ and the tangent to $\omega$ at $L$ ). This means that the points $K, G, E$, and $S$ are concyclic. Now, since $K S G T$ is a parallelogram, we obtain $\angle K E F=\angle K S G=180^{\circ}-\angle T K S=\angle A$. Since $K E=K F$, we also have $\angle K F E=\angle K E F=\angle A$.

After having proved this fact, one may finish the solution by different methods.

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Figure 1

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Figure 2

First method. We have just proved that all interesting triangles are similar to each other. This allows us to use the following lemma.



Lemma. Let $A B C$ be an arbitrary triangle. Choose two points $E_{1}$ and $E_{2}$ on the side $A C$, two points $F_{1}$ and $F_{2}$ on the side $A B$, and two points $K_{1}$ and $K_{2}$ on the side $B C$, in a way that the triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$ are similar. Then the six circumcircles of the triangles $A E_{i} F_{i}$, $B F_{i} K_{i}$, and $C E_{i} K_{i}(i=1,2)$ meet at a common point $Z$. Moreover, $Z$ is the centre of the spiral similarity that takes the triangle $E_{1} F_{1} K_{1}$ to the triangle $E_{2} F_{2} K_{2}$.

Proof. Firstly, notice that for each $i=1,2$, the circumcircles of the triangles $A E_{i} F_{i}, B F_{i} K_{i}$, and $C K_{i} E_{i}$ have a common point $Z_{i}$ by MiqueL's theorem. Moreover, we have

$\sphericalangle\left(Z_{i} F_{i}, Z_{i} E_{i}\right)=\sphericalangle(A B, C A), \quad \sphericalangle\left(Z_{i} K_{i}, Z_{i} F_{i}\right)=\sphericalangle(B C, A B), \quad \sphericalangle\left(Z_{i} E_{i}, Z_{i} K_{i}\right)=\sphericalangle(C A, B C)$.

This yields that the points $Z_{1}$ and $Z_{2}$ correspond to each other in similar triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$. Thus, if they coincide, then this common point is indeed the desired centre of a spiral similarity.

Finally, in order to show that $Z_{1}=Z_{2}$, one may notice that $\sphericalangle\left(A B, A Z_{1}\right)=\sphericalangle\left(E_{1} F_{1}, E_{1} Z_{1}\right)=$ $\sphericalangle\left(E_{2} F_{2}, E_{2} Z_{2}\right)=\sphericalangle\left(A B, A Z_{2}\right)$ (see Figure 2$)$. Similarly, one has $\sphericalangle\left(B C, B Z_{1}\right)=\sphericalangle\left(B C, B Z_{2}\right)$ and $\sphericalangle\left(C A, C Z_{1}\right)=\sphericalangle\left(C A, C Z_{2}\right)$. This yields $Z_{1}=Z_{2}$.

Now, let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $A C$ and $A B$, respectively, and let $R$ be the midpoint of $B C$ (see Figure 3). Then $R$ is the circumcentre of the cyclic quadrilateral $B C P Q$. Thus we obtain $\angle A P Q=\angle B$ and $\angle R P C=\angle C$, which yields $\angle Q P R=\angle A$. Similarly, we show that $\angle P Q R=\angle A$. Thus, all interesting triangles are similar to the triangle $P Q R$.

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Figure 3

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Figure 4

Denote now by $Z$ the common point of the circumcircles of $A P Q, B Q R$, and $C P R$. Let $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$ be two interesting triangles. By the lemma, $Z$ is the centre of any spiral similarity taking one of the triangles $E_{1} F_{1} K_{1}, E_{2} F_{2} K_{2}$, and $P Q R$ to some other of them. Therefore the triangles $Z E_{1} E_{2}$ and $Z F_{1} F_{2}$ are similar, as well as the triangles $Z E_{1} F_{1}$ and $Z P Q$. Hence

$$
\frac{E_{1} E_{2}}{F_{1} F_{2}}=\frac{Z E_{1}}{Z F_{1}}=\frac{Z P}{Z Q}
$$

Moreover, the equalities $\angle A Z Q=\angle A P Q=\angle A B C=180^{\circ}-\angle Q Z R$ show that the point $Z$ lies on the line $A R$ (see Figure 4). Therefore the triangles $A Z P$ and $A C R$ are similar, as well as the triangles $A Z Q$ and $A B R$. This yields

$$
\frac{Z P}{Z Q}=\frac{Z P}{R C} \cdot \frac{R B}{Z Q}=\frac{A Z}{A C} \cdot \frac{A B}{A Z}=\frac{A B}{A C},
$$

which completes the solution.



Second method. Now we will start from the fact that $\omega_{1}$ is tangent to the lines $K E$ and $K F$ (see Figure 5$)$. We prove that if $(E, F)$ is an interesting pair, then

$$
\frac{A E}{A B}+\frac{A F}{A C}=2 \cos \angle A
\tag{1}
$$

Let $Y$ be the intersection point of the segments $B E$ and $C F$. The points $B, K$, and $C$ are collinear, hence applying PASCAL's theorem to the degenerated hexagon $A F F Y E E$, we infer that $Y$ lies on the circle $\omega_{1}$.

Denote by $Z$ the second intersection point of the circumcircle of the triangle $B F Y$ with the line $B C$ (see Figure 6). By Miquel's theorem, the points $C, Z, Y$, and $E$ are concyclic. Therefore we obtain

$$
B F \cdot A B+C E \cdot A C=B Y \cdot B E+C Y \cdot C F=B Z \cdot B C+C Z \cdot B C=B C^{2}
$$

On the other hand, $B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos \angle A$, by the cosine law. Hence

$$
(A B-A F) \cdot A B+(A C-A E) \cdot A C=A B^{2}+A C^{2}-2 A B \cdot A C \cos \angle A
$$

which simplifies to the desired equality (1).

Let now $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ be two interesting pairs of points. Then we get

$$
\frac{A E_{1}}{A B}+\frac{A F_{1}}{A C}=\frac{A E_{2}}{A B}+\frac{A F_{2}}{A C}
$$

which gives the desired result.

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Figure 5

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Figure 6

Third method. Again, we make use of the fact that all interesting triangles are similar (and equi-oriented). Let us put the picture onto a complex plane such that $A$ is at the origin, and identify each point with the corresponding complex number.

Let $E F K$ be any interesting triangle. The equalities $\angle K E F=\angle K F E=\angle A$ yield that the ratio $\nu=\frac{K-E}{F-E}$ is the same for all interesting triangles. This in turn means that the numbers $E$, $F$, and $K$ satisfy the linear equation

$$
K=\mu E+\nu F, \quad \text { where } \quad \mu=1-\nu
\tag{2}
$$



Now let us choose the points $X$ and $Y$ on the rays $A B$ and $A C$, respectively, so that $\angle C X A=\angle A Y B=\angle A=\angle K E F$ (see Figure 7). Then each of the triangles $A X C$ and $Y A B$ is similar to any interesting triangle, which also means that

$$
C=\mu A+\nu X=\nu X \quad \text { and } \quad B=\mu Y+\nu A=\mu Y
\tag{3}
$$

Moreover, one has $X / Y=\overline{C / B}$.

Since the points $E, F$, and $K$ lie on $A C, A B$, and $B C$, respectively, one gets

$$
E=\rho Y, \quad F=\sigma X, \quad \text { and } \quad K=\lambda B+(1-\lambda) C
$$

for some real $\rho, \sigma$, and $\lambda$. In view of (3), the equation (2) now reads $\lambda B+(1-\lambda) C=K=$ $\mu E+\nu F=\rho B+\sigma C$, or

$$
(\lambda-\rho) B=(\sigma+\lambda-1) C .
$$

Since the nonzero complex numbers $B$ and $C$ have different arguments, the coefficients in the brackets vanish, so $\rho=\lambda$ and $\sigma=1-\lambda$. Therefore,

$$
\frac{E}{Y}+\frac{F}{X}=\rho+\sigma=1
\tag{4}
$$

Now, if $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ are two distinct interesting pairs, one may apply (4) to both pairs. Subtracting, we get

$$
\frac{E_{1}-E_{2}}{Y}=\frac{F_{2}-F_{1}}{X}, \quad \text { so } \quad \frac{E_{1}-E_{2}}{F_{2}-F_{1}}=\frac{Y}{X}=\frac{\bar{B}}{\bar{C}}
$$

Taking absolute values provides the required result.

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Figure 7
Let $(E, F)$ be an interesting pair. This time we prove that

$$
\frac{A M}{A K}=\cos \angle A
\tag{5}
$$

We introduce the circle $\omega$ passing through the points $K, S$, $A$, and $T$, together with the points $N$ and $L$ at which the line $A M$ intersect the line $S T$ and the circle $\omega$ for the second time, respectively. Let moreover $O$ be the centre of $\omega$ (see Figures 8 and 9). we note that $N$ is the midpoint of $S T$ and show that $K L \| S T$, which implies $\angle F A M=\angle E A K$.

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Figure 8

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Figure 9

Suppose now that $K \neq L$ (see Figure 8). Then $K L \| S T$, and consequently the lines $K M$ and $K L$ are perpendicular. It implies that the lines $L O$ and $K M$ meet at a point $X$ lying on the circle $\omega$. Since the lines $O N$ and $X M$ are both perpendicular to the line $S T$, they are parallel to each other, and hence $\angle L O N=\angle L X K=\angle M A K$. On the other hand, $\angle O L N=\angle M K A$, so we infer that triangles $N O L$ and $M A K$ are similar. This yields

$$
\frac{A M}{A K}=\frac{O N}{O L}=\frac{O N}{O T}=\cos \angle T O N=\cos \angle A
$$

If, on the other hand, $K=L$, then the points $A, M, N$, and $K$ lie on a common line, and this line is the perpendicular bisector of $S T$ (see Figure 9). This implies that $A K$ is a diameter of $\omega$, which yields $A M=2 O K-2 N K=2 O N$. So also in this case we obtain

$$
\frac{A M}{A K}=\frac{2 O N}{2 O T}=\cos \angle T O N=\cos \angle A
$$

Thus (5) is proved.

Let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $A C$ and $A B$, respectively (see Figure 10). We claim that the point $M$ lies on the line $P Q$. Consider now the composition of the dilatation with factor $\cos \angle A$ and centre $A$, and the reflection with respect to the angle bisector of $\angle B A C$. This transformation is a similarity that takes $B, C$, and $K$ to $P, Q$, and $M$, respectively. Since $K$ lies on the line $B C$, the point $M$ lies on the line $P Q$.



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Figure 10

Suppose that $E \neq P$. Then also $F \neq Q$, and by Menelaus' theorem, we obtain

$$
\frac{A Q}{F Q} \cdot \frac{F M}{E M} \cdot \frac{E P}{A P}=1
$$

Using the similarity of the triangles $A P Q$ and $A B C$, we infer that

$$
\frac{E P}{F Q}=\frac{A P}{A Q}=\frac{A B}{A C}, \quad \text { and hence } \quad \frac{E P}{A B}=\frac{F Q}{A C}
$$

The last equality holds obviously also in case $E=P$, because then $F=Q$. Moreover, since the line $P Q$ intersects the segment $E F$, we infer that the point $E$ lies on the segment $A P$ if and only if the point $F$ lies outside of the segment $A Q$.

Let now $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ be two interesting pairs. Then we obtain

$$
\frac{E_{1} P}{A B}=\frac{F_{1} Q}{A C} \quad \text { and } \quad \frac{E_{2} P}{A B}=\frac{F_{2} Q}{A C}
$$

If $P$ lies between the points $E_{1}$ and $E_{2}$, we add the equalities above, otherwise we subtract them. In any case we obtain

$$
\frac{E_{1} E_{2}}{A B}=\frac{F_{1} F_{2}}{A C}
$$

which completes the solution.