The plane $\alpha$ intersects the edges $AB$, $BC$, $CD$ and $DA$ of the tetrahedron $ABCD$ at points $X, Y, Z$ and $T$, respectively. It turned out, that points $Y$ and $T$ lie on a circle $\omega$ constructed with segment $XZ$ as the diameter. Point $P$ is marked in the plane $\alpha$ so that the lines $P Y$ and $P T$ are tangent to the circle $\omega$.Prove that the midpoints of the edges are $AB$, $BC$, $CD,$ $DA$ and the point $P$ lie in the same plane.
Problem
Source: All-Russian MO 2023 Final stage 11.6
Tags: geometry, 3D geometry, tetrahedron
rms
26.08.2024 19:40
parmenides51 wrote: The plane $\alpha$ intersects the edges $AB$, $BC$, $CD$ and $DA$ of the tetrahedron $ABCD$ at points $X, Y, Z$ and $T$, respectively. It turned out, that points $Y$ and $T$ lie on a circle $\omega$ constructed with segment $XZ$ as the diameter. Point $P$ is marked in the plane $\alpha$ so that the lines $P Y$ and $P T$ are tangent to the circle $\omega$.Prove that the midpoints of the edges are $AB$, $BC$, $CD,$ $DA$ and the point $P$ lie in the same plane. From here:
**Solution.** From the conditions of the problem, we immediately obtain that \(\angle XYZ = 90^\circ = \angle XTZ\). Let \(Q\) be the intersection point of lines \(XY\) and \(ZT\), and \(R\) be the intersection point of lines \(ZY\) and \(XT\) (see Fig. 3). Without loss of generality, we can assume that point \(Z\) lies on segments \(RY\) and \(QT\). Since point \(R\) lies in planes \(ABD\) and \(BCD\), it lies on line \(BD\). Similarly, point \(Q\) lies on line \(AC\).
Notice that \(RY\) and \(QT\) are altitudes of triangle \(XQR\). Then \(Z\) is the intersection point of the altitudes of this triangle.
Thus, \(XZ \perp QR\). Let \(M\) be the midpoint of segment \(QR\). Since \(\angle QYR = 90^\circ\), \(YM = MR = RQ\) by the property of the median in a right triangle. Hence, \(\angle MYR = \angle YRQ = 90^\circ - \angle XQR = \angle ZXQ\). Therefore, line \(YM\) touches the circumcircle \(\omega\). Similarly, line \(TM\) also touches the circumcircle \(\omega\), so points \(M\) and \(P\) coincide.
Consider two parallel planes \(\beta\) and \(\gamma\), one of which contains segment \(AC\), and the other—segment \(BD\). Note that the midpoints of all segments connecting a point from plane \(\beta\) and a point from plane \(\gamma\), lie in one plane parallel to \(\beta\) and \(\gamma\). Indeed, if we introduce Cartesian coordinates so that one of these planes is given by the equation \(z = 0\), and the other by \(z = h\) (where \(h\) is the distance between planes \(\beta\) and \(\gamma\)), then the midpoints of the considered segments lie in the plane \(z = h/2\). Applying this observation to segments \(AB\), \(BC\), \(CD\), \(DA\), and \(QR\), we find that their midpoints lie on one plane, which was required to be proven.
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