Резюме. The International Mathematical Olympiad is one of the respectable events and one of the most long-lived international educational and scientific competitions. It is the largest, oldest and most prestigious scientific Olympiad for high school students. The \(59{ }^{\text {th }}\) edition of the event took place in Cluj-Napoca, Romania, \(3-14\) July 2018. The present paper is dedicated to the sixth problem on the Olympiad paper. A detailed analysis of the problem is proposed in a methodological way, which will be useful for students and teachers in the preparatory process for future participations in mathematical competitions.

Ключови думи: Olympiad; problem solving

The problem 6 on the paper of the \(59{ }^{\text {th }}\) International mathematical Olympiad was solved fully (7 points) by 18 participants, 5 students were marked with 6 points, 2 with 5 points, 5 with 4 points, 11 with 3 points, 26 with 2 points, 108 with 1 point and 419 with 0 points. The mean result of all the 594 participants in the Olympiad from 107 countries is 0. 638, which shows that the problem is hard and needs a detailed analysis.

Problem 6. A convex quadrilateral \(A B C D\) satisfies \(A B . C D=B C . D A\). . Point \(X\) lies inside \(A B C D\) so that \(\angle X A B=\angle X C D\) and \(\angle X B C=\angle X D A\). Prove that \(\angle B X A+\angle D X C=180^{\circ}\).

Lemma 1. Each convex quadrilateral has a unique interior point \(X\) such that \(\angle X A B=\angle X C D\) and \(\angle X B C=\angle X D A\).

Proof: The following cases are possible:

1) If \(A B C D\) is a parallelogram, then the point \(X\) is the intersection point of the diagonals. This follows from the equality of the cross-opposite angles of the parallelogram. Reversely, the equality of the mentioned angles implies that the point \(X\) should belong to each of the diagonals and consequently it belongs to both the diagonals simultaneously. Thus the point \(X\) is unique.

2.1) If \(A B C D\) is a trapezoid with \(A B \| C D\) and \(B C \cap A D=W\), then the common point of the diagonal \(A C\) and the circumcircle \(k(B D W)\) of \(\triangle B D W\) is the desired point \(X\). The proof is the following:

\[ \begin{aligned} & \text { i) } \angle X A B=\angle X C D \text { (cross-opposite angles); } \\ & \text { ii) } \angle X B C=\tfrac{\overparen{X D W}}{2}=\tfrac{\overparen{X D}+\overparen{D W}}{2}=\angle X D A \text {. } \end{aligned} \]

Reversely, the equality \(\angle X A B=\angle X C D\) implies, that the point \(X\) belongs to the diagonal \(A C\), while the equality \(\angle X B C=\angle X D A\) is verified when \(X\) is on the circle \(k(B D W)\). Since \(A C\) and \(k(B D W)\) have only one common point, which is in the interior of \(A B C D\), then the point \(X\) is unique. The second intersection point \(Y\) of the line \(A C\) and \(k(B D W)\) is such, that \(\angle Y A B=180^{\circ}-\angle Y C D\) and \(\angle Y B C=180^{\circ}-\angle Y D A\).

2.2) If \(A B C D\) is a trapezoid with \(B C \| A D\) and \(A B \cap C D=V\), then the common point of the diagonal \(B D\) and the circumcircle \(k(A C V)\) of \(\triangle A C V\) is the desired point \(X\). The proof is the following:

i) \(\angle X A B=\tfrac{\overparen{X C V}}{2}=\tfrac{\overparen{X C}+\overparen{C V}}{2}=\angle X C D\);

ii) \(\angle X B C=\angle X D A\) (cross-opposite angles).

The uniqueness of the point \(X\) could be proved in the same manner as in the previous case. The second common point \(Y\) of the line \(B D\) and \(k(A C V)\) is such, that \(\angle Y A B=180^{\circ}-\angle Y C D\) and \(\angle Y B C=180^{\circ}-\angle Y D A\).

3) If \(A B C D\) is without parallel sides and \(A B \cap C D=V, B C \cap A D=W\), then the interior point of \(A B C D\), which is the intersection point of the circumcircles \(k(A C V)\) and \(k(B D W)\) of \(\triangle A C V\) and \(\triangle B D W\), respectively, is the desired point \(X\). The proof is the following:

i) \(\angle X A B=\tfrac{\overparen{X C V}}{2}=\tfrac{\overparen{X C}+\overparen{C V}}{2}=\angle X C D\);

ii) \(\angle X B C=\tfrac{\overparen{X D W}}{2}=\tfrac{\overparen{X D}+\overparen{D W}}{2}=\angle X D A\).

Reversely, the equality \(\angle X A B=\angle X C D\) means, that the point \(X\) belongs to the circle \(k(A C V)\), while the equality \(\angle X B C=\angle X D A\) is verified when \(X\) is on the circle \(k(B D W)\). Since \(A C\) and \(k(B D W)\) have only one common point, which is in the interior of \(A B C D\), then the point \(X\) is unique. The second intersection point \(Y\) of \(k(A C V)\) and \(k(B D W)\) is such, that \(\angle Y A B=180^{\circ}-\angle Y C D\) and \(\angle Y B C=180^{\circ}-\angle Y D A\).

Lemma 2. If the sides of the convex quadrilateral \(A B C D\) satisfy the equality \(A B . C D=B C . D A\), then:

а) the angular bisectors of the angles \(A B C\) and \(C D A\) meet in the point \(I\) from the diagonal \(A C\);

b) the angular bisectors of the angles \(B A D\) and \(D C B\) meet in the point \(I^{\prime}\)

from the diagonalon the diagonalProof: Rewrite the \(B D\) \(A C\), that given. \(\tfrac{I A}{I C}=\tfrac{B A}{B C}\) equality in, then the form \(B I\) is the angular bisector of \(\tfrac{B A}{B C}=\tfrac{D A}{D C}\). If \(I\) is such a point \(\angle A B C\). On the other hand, the equality \(\tfrac{I A}{I C}=\tfrac{D A}{D C}\) is true, which means, that \(D I\) is the angular bisector of \(\angle C D A\). The assertion b) could be obtained analogously.

In fact, if \(B A \neq B C\), then there exists a point \(J\) on the line \(B C\) with the property \(\tfrac{J A}{J C}=\tfrac{B A}{B C}\). The circle \(\omega\) with diameter \(I J\) is the locus of the points \(M\), for which \(\tfrac{M A}{M C}=\tfrac{B A}{B C}\). The circle \(\omega\) is known to be the Apollonius circle for the segment \(A C\) under the ratio \(\tfrac{B A}{B C}\). For this reason, if an arbitrary \(\triangle A B C\) is given, then the point \(D\) from the Apollonius circle \(\omega\) is the fourth vertex of the quadrilateral \(A B C D\), for which \(A B . C D=B C . D A\).CD = BC.DA . In such a way we come upon an idea for the construction of a quadrilateral satisfying the conditions of the problem under consideration. On the other hand, lemma 1 gives a possibility to construct the point \(X\) for this quadrilateral.

Lemma 3. Let \(A B C\) be an arbitrary triangle, while \(B_{1}\) and \(B_{2}\) be such points on the line \(A C\), that the lines \(B B_{1}\) and \(B B_{2}\) are symmetric with respect to the angular bisector of \(\angle A B C\). Then \(\tfrac{\overline{C B_{1}}}{\overline{A B_{1}}} \cdot \tfrac{\overline{C B_{2}}}{\overline{A B_{2}}}=\tfrac{B C^{2}}{A B^{2}}\).

Proof: For the areas ofthe triangles \(B C B_{1}\) and \(B A B_{2}\) we have, that \(\tfrac{S_{B C B_{1}}}{S_{B A B_{2}}}=\tfrac{C B_{1}}{A B_{2}}\) and \(\tfrac{S_{B C B_{1}}}{S_{B A B_{2}}}=\tfrac{B C \cdot B B_{1} \cdot \sin \angle B_{1} B C}{A B \cdot B B_{2} \cdot \sin \angle B_{2} B A}=\tfrac{B C \cdot B B_{1}}{A B \cdot B B_{2}}\). Consequently \(\tfrac{C B_{1}}{A B_{2}}=\tfrac{B C \cdot B B_{1}}{A B \cdot B B_{2}}\). Analogously, using the areas of the triangles \(C B B_{2}\) and \(A B B_{1}\) we get the equality \(\tfrac{C B_{2}}{A B_{1}}=\tfrac{B C . B B_{2}}{A B . B B_{1}}\). Multiplying the obtained equalities we find, that \(\tfrac{C B_{1}}{A B_{1}} \cdot \tfrac{C B_{2}}{A B_{2}}=\tfrac{B C^{2}}{A B^{2}}\).

Lemma 4. If the convex quadrilateral \(A B C D\) satisfies the equality \(A B . C D=B C . D A\), then:

а) the symmetric images of the diagonal \(B D\) with respect to the angular bisectors of \(\angle A B C\) and \(\angle C D A\) meet in a point \(K\) from the diagonal \(A C\);

b) the symmetric images of the diagonal \(A C\) with respect to the angular bisectors of \(\angle B A D\) and \(\angle D C B\) meet in a point \(K^{\prime}\) from the diagonal \(B D\).

Proof: Denote by \(U\) the common point of the diagonals \(A C\) and \(B D\). Let the symmetric images of \(B D\) with respect to the angular bisectors of \(\angle A B C\) and \(\angle C D A\) meet \(A C\) in the points \(K_{1}\) and \(K_{2}\), respectively. Applying lemma 3 to \(\triangle A B C\) we obtain the equality \(\tfrac{C U}{A U} \cdot \tfrac{C K_{1}}{A K_{1}}=\tfrac{B C^{2}}{A B^{2}}\). Applying lemma 3 to \(\triangle C D A\) we obtain \(\tfrac{A U}{C U} \cdot \tfrac{A K_{2}}{C K_{2}}=\tfrac{D A^{2}}{C D^{2}}\). Multiplying the two equalities we deduce, that \(\tfrac{C K_{1} \cdot A K_{2}}{A K_{1} \cdot C K_{2}}=\left(\tfrac{B C \cdot D A}{A B \cdot C D}\right)^{2}=1\). Therefore, \(\tfrac{A K_{2}}{C K_{2}}=\tfrac{A K_{1}}{C K_{1}}\). Since the points \(K_{1}\) and \(K_{2}\) belong to the segment \(A C\), the last equality means, that the points \(K_{1}\) and \(K_{2}\) coincide. The assertion for the point \(K^{\prime}\) could be proven analogously.

It follows from lemma 4 that the point \(I\) is the center of the incircle of \(\triangle B D K\). Thus, \(A C\) is the angular bisector of \(\angle B K D\). Consequently \(180^{\circ}=\angle A K D+\angle D K C=\angle A K D+\angle B K C\). Thus we obtain the following

Conclusion. The equalities \(\angle A K D+\angle B K C=180^{\circ}\) and \(\angle B K^{\prime} A+\angle C K^{\prime} D=180^{\circ}\) are satisfied.

Lemma 5. If the convex quadrilateral \(A B C D\) satisfies the equality \(A B . C D=B C . D A\), then:

а) The second common point \(X\) of the circumcircles \(k(B C K)\) and \(k(D A K)\) of \(\triangle B C K\) and \(\triangle D A K\), respectively, is such that \(\angle X A B=\angle X C D\) and \(\angle X B C=\angle X D A\).

b) The second common point \(X\) of the circumcircles \(k\left(B C K^{\prime}\right)\) and \(k\left(D A K^{\prime}\right)\) of \(\triangle B C K^{\prime}\) and \(\triangle D A K^{\prime}\), respectively, is such that \(\angle X A B=\angle X C D\) and \(\angle X B C=\angle X D A\).

Proof: We will consider the first case only. Let \(A B \cap A D=V\) and the second common point of \(k(B C K)\) and \(k(D A K)\) be \(X\). In order to prove the first equality it is enough to establish, that the point \(X\) belongs to the circumcircle \(k(A C V)\) of \(\triangle A C V\) (as shown in lemma 1). The following equalities are true:

\[ \begin{gathered} \angle C X A=\angle C X K+\angle K X A=180^{\circ}-\angle C B K+\tfrac{\overparen{A K}}{2}=180^{\circ}-\angle D B A+\angle A D K= \\ =\angle D B V+\angle B D V=180^{\circ}-\angle A V D \end{gathered} \]

Consequaently \(\angle C X A+\angle A V D=180^{\circ}\). This equality means that the point \(X\) belongs to the circumcircle of \(\triangle A C V\). For this reason \(\angle X A B=\angle X C D\). The second equality could be obtained in the following way:

\(\angle X B C=\tfrac{\overparen{X C}}{2}=\angle X K C=180^{\circ}-\angle A K X=\angle X D A\).

The assertion for the other couple of circles under the condition \(A D \cap B C=W\) could be proven in the same way.

If \(A B \| C D\) and the second common point of \(k(B C K)\) and \(k(D A K)\) is \(X\), as in the case just considered, we obtain \(\angle C X A=180^{\circ}-\angle D B A+\angle B D C=180^{\circ}\). This means that the points \(C, X\) and \(A\) are collinear, i.e. the point \(X\) is on the diagonal \(A C\). Consequently, the common points \(X\) and \(K\) of the circles \(k(B C K)\) and \(k(D A K)\) are on the diagonal \(A C\). But the circle \(k(B C K)\) intersects the line \(A C\) in the point \(C\). Since a circle and a line could have no more than two common points, then the point \(X\) coincides with \(K\). Therefore, in this case the circles \(k(B C K)\) and \(k(D A K)\) are tangent at the point \(X\).

Analogously, if \(A D \| B C\), the circles \(k\left(B C K^{\prime}\right)\) and \(k\left(D A K^{\prime}\right)\) are tangent at the point \(X \equiv K^{\prime}\).

Solution of problem 6: If \(A B C D\) is a parallelogram, then the equality \(A B . C D=B C . D A\) is satisfied only in the case when \(A B C D\) is a rhombus. Since the point \(X\) is the intersection point of its diagonals, then \(\angle B X A+\angle D X C=90^{\circ}+90^{\circ}=180^{\circ}\). Let now \(A B C D\) is such, that two of its sides at least are not parallel to each other. For definiteness let \(A B \cap C D=V\). Since (according to lemma 5) the point \(X\) belongs to the circles \(k(B C K)\) and \(k(D A K)\), then \(\angle D X A+\angle B X C=\angle A K D+\angle B K C\). From the conclusion we obtain, that \(\angle D X A+\angle B X C=180^{\circ}\). But \((\angle B X A+\angle D X C)+(\angle D X A+\angle B X C)=360^{\circ}\). Therefore \(\angle B X A+\angle D X C=180^{\circ}\). If we consider the point \(X\) as a common point of the circles \(k\left(B C K^{\prime}\right)\) and \(k\left(D A K^{\prime}\right)\), we deduce the desired equality as a consequence of lemma 5 and the conclusion in the following way:

\[ \angle B X A+\angle D X C=\angle B K^{\prime} A+\angle C K^{\prime} D=180^{\circ} . \]

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