Образователни технологии
A TRIANGLE AND A TRAPEZOID WITH A COMMON CONIC
Резюме. The aim of the present note is to propose a generalization of Problem 1 on the IMO’2018 paper. The International Mathematical Olympiad (IMO) is the most prestigious scientific Olympiad for high school students. Its \(59{ }^{\text {th }}\) edition took place in Cluj-Napoca, Romania, 3–14 July 2018. The problem 1 on the paper was solved fully (7 points) by 381 participants, 7 students were marked with 6 points, 7 with 5 points, 10 with 4 points, 15 with 3 points, 24 with 2 points, 54 with 1 point and 96 with 0 points. The mean result of all the 594 participants in the Olympiad from 107 countries is 4, 934, which shows that the problem is easy and has not bordered most of the contestants. Nevertheless it turns out to be interesting and originates rich in content ideas.
Ключови думи: Olympiad; problem solving; triangle; trapezoid; conic
The problem under consideration is the following:
Let \(\Gamma\) be the circumcircle of an acute triangle \(A B C\), while \(D\) and \(E\) be on the sides \(A B\) and \(A C\) respectively verifying \(A D=A E\). If the perpendicular bisectors of the segments \(B D\) and \(C E\) meet the small \(\operatorname{arcs} \overparen{A B}\) and \(\overparen{A C}\) of \(\Gamma\) in points \(F\) and \(G\), respectively, prove that the lines \(D E\) and \(F G\) are parallel or coinciding.
First, replace the circumcircle \(\Gamma\) of \(\triangle A B C\) by an arbitrary conic \(k\) with center \(O\). Let \(D\) be an arbitrary point on the line \(A B\). For the identification of point \(E\) on the line \(A C\), we need an interpretation of the relation \(A D=A E\) from the initial problem. The relation shows that \(\triangle D E A\) is isosceles and that the angular bisector of \(∢ B A C\) is the perpendicular bisector of \(D E\). In other words, the angular bisector passes through the mid-point of \(D E\) and is conjugated with \(D E\) with respect to \(\Gamma\). On the other hand, the angular bisector in question crosses the diameter of \(\Gamma\) through the midpoint of \(B C\) in a point on \(\Gamma\). For this reason we consider the following construction: The diameter of \(k\) through the midpoint of the side \(B C\) meets \(k\) in point \(N\). The line through \(D\), which is conjugate with \(A N\), meets \(A C\) in point \(E\). Construct the lines \(p_{c}\) and \(p_{b}\) through the midpoints of \(B D\) and \(C E\), respectively, to be conjugated with the lines \(A B\) and \(A C\). Let the lines \(p_{c}\) and \(p_{b}\) meet \(k\) in the point couples \(\left(F_{1}, F_{2}\right)\) and \(\left(G_{1}, G_{2}\right)\), , respectively
The following properties of the above construction are satisfied:
А) Two of the lines \(F_{1} G_{1}, F_{2} G_{2}, F_{1} G_{2}\) and \(F_{2} G_{1}\) are parallel to \(D E\) or one of them coincides with \(D E\);
B) If \(F_{1} G_{2} \cap F_{2} G_{1}=P\) and \(p_{c} \cap p_{b}=M\), then the points \(O, M\) and \(P\) are on a line which is parallel to \(A N\).
For the proof of the first assertion we will use barycentric coordinates with respect to \(\triangle A B C\) with \(A(1,0,0), B(0,1,0), C(0,0,1), O\left(x_{0}, y_{0}, z_{0}\right)\left(x_{0}+y_{0}+z_{0}=1\right)\) and \(D(\delta, 1-\delta, 0)\) ( (see [1]).
The equation of \(k\) is the following
(1)\(k:\left(1-2 x_{0}\right) x_{0} y z+\left(1-2 y_{0}\right) y_{0} z x+\left(1-2 z_{0}\right) z_{0} x y=0\).
By this equation and the parametric equations
\[ x=-2 x_{0} t, y=\tfrac{1}{2}+\left(1-2 y_{0}\right) t, z=\tfrac{1}{2}+\left(1-2 z_{0}\right) t \] of the diameter of \(k\) through the midpoint of \(B C\) we find the coordinates of the intersection points \(N_{1}\) and \(N_{2}\) in the following form
(2) \(N_{1}\left(\tfrac{\left(1-2 y_{0}\right)\left(1-2 z_{0}\right)-2 D_{0}}{\left(1-2 y_{0}\right)\left(1-2 z_{0}\right)} x_{0}, \tfrac{y_{0}\left(1-2 z_{0}\right)+D_{0}}{1-2 z_{0}}, \tfrac{z_{0}\left(1-2 y_{0}\right)+D_{0}}{1-2 y_{0}}\right)\),
(3) \(N_{2}\left(\tfrac{\left(1-2 y_{0}\right)\left(1-2 z_{0}\right)+2 D_{0}}{\left(1-2 y_{0}\right)\left(1-2 z_{0}\right)} x_{0}, \tfrac{y_{0}\left(1-2 z_{0}\right)-D_{0}}{1-2 z_{0}}, \tfrac{z_{0}\left(1-2 y_{0}\right)-D_{0}}{1-2 y_{0}}\right)\), където \(D_{0}=\sqrt{y_{0} z_{0}\left(1-2 y_{0}\right)\left(1-2 z_{0}\right)}\).
We assume that \(N \equiv N_{1}\). The other case could be considered analogously.
Further it will be necessary to determine the coordinates of a vector, which is conjugate with a given vector. It follows from the results in [2], that in case the vector \(\overrightarrow{v}\left(v_{1}, v_{2}, v_{3}\right)\) is conjugate with the vector \(\overrightarrow{u}\left(u_{1}, u_{2}, u_{3}\right)\), then the following equations are satisfied:
(4)\[ \begin{aligned} & v_{1}=\left(1-2 x_{0}\right)\left[\left(y_{0}-z_{0}\right) u_{1}-x_{0} u_{2}+x_{0} u_{3}\right] \\ & v_{2}=\left(1-2 y_{0}\right)\left[y_{0} u_{1}+\left(z_{0}-x_{0}\right) u_{2}-y_{0} u_{3}\right] \\ & v_{3}=\left(1-2 z_{0}\right)\left[-z_{0} u_{1}+z_{0} u_{2}+\left(x_{0}-y_{0}\right) u_{3}\right] \end{aligned} \]
A vector which is collinear with \(\overrightarrow{A N}\) has the following coordinates:
\[ \begin{aligned} & -\left(1-2 y_{0}\right)\left(1-2 z_{0}\right)\left(y_{0}+z_{0}\right)-2 x_{0} D_{0} \\ & \left(1-2 y_{0}\right)\left(1-2 z_{0}\right) y_{0}+\left(1-2 y_{0}\right) D_{0} \\ & \left(1-2 y_{0}\right)\left(1-2 z_{0}\right) z_{0}+\left(1-2 z_{0}\right) D_{0} \end{aligned} \]
Applying these coordinares and the formulae (4) we deduce that the vector \[ \left(\left(1-2 x_{0}\right)\left(z_{0}-y_{0}\right), 2 y_{0}^{2}-y_{0}-D_{0},-2 z_{0}^{2}+z_{0}+D_{0}\right) \] is conjugate with \(\overrightarrow{A N}\). Consequently the line \(d\), which passes through the point \(D\) and is conjugate with the line \(A N\) has the following parametric equations:
\[ d: x=\delta+\left(1-2 x_{0}\right)\left(z_{0}-y_{0}\right) d_{0}, y=1-\delta+\left(2 y_{0}^{2}-y_{0}-D_{0}\right) d_{0}, z=\left(-2 z_{0}^{2}+z_{0}+D_{0}\right) d_{0} \] From the last equations we obtain the coordinates of the common point \(E\) of \(d\) and the line \(A C: y=0\) :
\[ E\left(-\tfrac{\left[\left(1-2 z_{0}\right) z_{0}+D_{0}\right] \delta+\left(1-2 x_{0}\right)\left(z_{0}-y_{0}\right)}{2 y_{0}^{2}-y_{0}-D_{0}}, 0, \tfrac{\left(-2 z_{0}^{2}+z_{0}+D_{0}\right)(\delta-1)}{2 y_{0}^{2}-y_{0}-D_{0}}\right) \] Further, by the formulae (4) we determine the conjugate directions of the vectors \(\overrightarrow{B A}(1,-1,0)\) and \(\overrightarrow{A C}(-1,0,1)\). Thus, we obtain the parametric equations of the lines \(p_{c}\) and \(p_{b}\) :
\[ \begin{gathered} p_{c}:\left\{\begin{array}{l} x=\tfrac{\delta}{2}+\left(1-2 x_{0}\right) t_{c} \\ y=\tfrac{2-\delta}{2}+\left(1-2 y_{0}\right) t_{c} \\ z=-2 z_{0} t_{c} \end{array}\right. \\ p_{b}:\left\{\begin{array}{l} x=-\tfrac{\left[\left(1-2 z_{0}\right) z_{0}+D_{0}\right] \delta+\left(1-2 x_{0}\right)\left(z_{0}-y_{0}\right)}{2\left(2 y_{0}^{2}-y_{0}-D_{0}\right)}+\left(1-2 x_{0}\right) t_{b} \\ y=-2 y_{0} t_{b} \\ z=\tfrac{\left(-2 z_{0}^{2}+z_{0}+D_{0}\right) \delta+2\left(y_{0}^{2}+z_{0}^{2}-D_{0}\right)+x_{0}-1}{2\left(2 y_{0}^{2}-y_{0}-D_{0}\right)}+\left(1-2 z_{0}\right) t_{b} \end{array}\right. \end{gathered} \]
From these equations and (1) we find the coordinates of the points \(F_{1}, F_{2}, G_{1}\) and \(G_{2}\) in the form:
\[ \begin{aligned} & F_{1}\left(\tfrac{\delta}{2}-\tfrac{\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)+D_{c}}{2\left(1-2 y_{0}\right)}, \tfrac{2-\delta}{2}-\tfrac{\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)+D_{c}}{2\left(1-2 x_{0}\right)}, \tfrac{\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)+D_{c}}{\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)} z_{0}\right) \\ & F_{2}\left(\tfrac{\delta}{2}-\tfrac{\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)-D_{c}}{2\left(1-2 y_{0}\right)}, \tfrac{2-\delta}{2}-\tfrac{\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)-D_{c}}{2\left(1-2 x_{0}\right)}, \tfrac{\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)-D_{c}}{\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)} z_{0}\right) \end{aligned} \] \[ \begin{gathered} x_{G_{1}}=-\tfrac{\left[\left(1-2 z_{0}\right) z_{0}+D_{0}\right] \delta+\left(1-2 x_{0}\right)\left(z_{0}-y_{0}\right)}{2\left(2 y_{0}^{2}-y_{0}-D_{0}\right)}-\tfrac{\left(1-2 z_{0}\right)\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)+D_{b}}{2\left(1-2 z_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)}, \\ y_{G_{1}}=\tfrac{\left(1-2 z_{0}\right)\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)+D_{b}}{\left(1-2 z_{0}\right)\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)}, \\ z_{G_{1}}=\tfrac{\left(-2 z_{0}^{2}+z_{0}+D_{0}\right) \delta+2\left(y_{0}^{2}+z_{0}^{2}-D_{0}\right)+x_{0}-1}{2\left(2 y_{0}^{2}-y_{0}-D_{0}\right)}-\tfrac{\left(1-2 z_{0}\right)\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)+D_{b}}{2\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)}, \\ x_{G_{2}}=-\tfrac{\left[\left(1-2 z_{0}\right) z_{0}+D_{0}\right] \delta+\left(1-2 x_{0}\right)\left(z_{0}-y_{0}\right)}{2\left(2 y_{0}^{2}-y_{0}-D_{0}\right)}-\tfrac{\left(1-2 z_{0}\right)\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)-D_{b}}{2\left(1-2 z_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)}, \\ y_{G_{2}}=\tfrac{\left(1-2 z_{0}\right)\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)-D_{b}}{\left(1-2 z_{0}\right)\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)}, \\ z_{G_{2}}=\tfrac{\left(-2 z_{0}^{2}+z_{0}+D_{0}\right) \delta+2\left(y_{0}^{2}+z_{0}^{2}-D_{0}\right)+x_{0}-1}{2\left(2 y_{0}^{2}-y_{0}-D_{0}\right)}-\tfrac{\left(1-2 z_{0}\right)\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)-D_{b}}{2\left(1-2 x_{0}\right)\left(-2 y_{0}^{2}+y_{0}+D_{0}\right)}, \end{gathered} \] where
\(D_{c}=\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)\left[-\left(1-2 z_{0}\right) \delta^{2}+2\left(1-2 z_{0}\right) \delta+\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)\right]\) and \(D_{b}=\left[z_{0}\left(1-2 z_{0}\right)+D_{0}\right] D_{c}\).
From the coordinates of \(F_{1}\) and \(F_{2}\) we obtain the equations
\[ \begin{gathered} x_{G_{1}}-x_{F_{1}}=\left(z_{0}-y_{0}\right)\left(1-2 x_{0}\right) m, y_{G_{1}}-y_{F_{1}}=\left(2 y_{0}^{2}-y_{0}-D_{0}\right) m \\ z_{G_{1}}-z_{F_{1}}=\left(-2 z_{0}^{2}+z_{0}+D_{0}\right) m \end{gathered} \] where \(m=\tfrac{\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)\left(1-2 z_{0}\right)(\delta-1)+\left[\left(1-2 y_{0}\right)\left(1-2 z_{0}\right)+2 D_{0}\right] D_{c}}{2\left(1-2 x_{0}\right)\left(1-2 y_{0}\right)\left(1-2 z_{0}\right)\left(2 y_{0}^{2}-y_{0}-D_{0}\right)}\).
On the other hand \(\overrightarrow{D E}\left(\left(z_{0}-y_{0}\right)\left(1-2 x_{0}\right) n,\left(2 y_{0}^{2}-y_{0}-D_{0}\right) n,\left(-2 z_{0}^{2}+z_{0}+D_{0}\right) n\right)\), where \(n=\tfrac{1-\delta}{2 y_{0}^{2}-y_{0}-D_{0}}\). Consequently \(F_{1} G_{1} \| D E\) or \(F_{1} G_{1} \equiv D E\). Analogously it could be shown that \(F_{2} G_{2} \| D E\) or \(F_{2} G_{2} \equiv D E\).
The second assertion from the beginning could be obtained in the following way: Since \(M\) and \(P\) are the intersection points of the diagonals and the leg-lines of the trapezoid \(G_{1} F_{1} G_{2} F_{2}\), it follows from Steiners’s's theorem [3] that the line \(M P\) passes through the mid-points of the bases \(G_{1} F_{1}\) and \(G_{2} F_{2}\). Consequently, the line \(M P\) is a diameter of \(k\) and for this reason it passes through its center \(O\). Additionally the line \(M P\) is conjugate with the bases of the trapezoid \(G_{1} F_{1} G_{2} F_{2}\) and the line \(D E\). But the last is conjugate with \(A N\) according to the construction. Consequently \(M P \| A N\).
REFERENCES
Paskalev, G. & I. Chobanov (1985). Notable points in the triangle. Sofia: Narodna Prosveta [Паскалев, Г. & И. Чобанов (1985). Забележителни точки в триъгълника. София: Народна просвета.]
Grozdev, S. & V. Nenkov (2015). Geometric construction of Cheva curve, Mathematics and Informatics, 1, 52 – 57. [Гроздев, С. & В. Ненков (2015). Геометрична конструкция на крива на Чева, Математика и информатика, 1, 52 – 57.]
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