Реши задачу. Задача 10-го класса Реши, для построе...

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प्रश्न

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Реши задачу. Задача 10-го класса
Реши, для построения сечения, если возможно и надо будет, используй параллельный перенос. ЭТО ЗАЧАДА ПО ГЕОМЕТРИИ

विचार

Breaking down problem statement

The problem is in Russian and asks about the parallelism of a line and a plane. It’s about a tetrahedron with vertices D, A, B, and C. Points K, M, and N divide specific edges of the tetrahedron according to given ratios. The task involves determining the length of segment AP when a plane intersects edge AB at P.

Approaching the intersection problem

We need to determine where the plane KMN intersects the line AB. The ratio of intersection along AB depends on the plane defined by points K, M, and N. This calls for using a technique like mass points, the section theorem, or barycentric coordinates. We'll choose coordinates for the tetrahedron to calculate the ratio AP/AB.

Considering intersection with ABD

I’m thinking about another method: what if we consider the intersection of the plane with another plane, like ABD? Since AB is an edge in the ABD face, the plane intersects ABD along a line through K. But M and N aren't in ABD—however, the line will still pass through K, which is on AD. This will also intersect BD because M lies on DC.

Refining the intersection approach

The plane intersects ABD along a line through K and some point on BD. To find this line, I might need to check the intersection with another edge of ABD, like DB, possibly using parallel transfer. Since M and N lie on DC and BC, projecting from C along lines parallel to MN might help too.

Considering Menelaus' Theorem

I’m thinking of using Menelaus' theorem to solve this—if a plane intersects three edges of a tetrahedron that don't all meet at the same vertex, the ratios along a fourth edge can be determined. The plane cuts edges AD, DC, and BC, and I'll use linear interpolation and projections to determine where the plane intersects AB.

Exploring plane intersections

Okay, I’m working through the geometry of planes in 3D. Menelaus' Theorem tells us that in triangle DCB, if M and N divide the sides proportionally, the line MN is parallel to DB. This helps tell us that when a plane intersects ABD along KD, it’s parallel to BD due to the alignment of MN and DB.

Verifying the intersection line

I'm working through the geometry of the planes and confirming that the intersection line between planes π and ABD is unique. It’s parallel to DB based on properties of parallel lines in 3D. Then by using the intercept theorem, I calculate that the point P divides AB in a ratio and find that AP = 8.

Thought for 2m 37s

उत्तर

Ответ: AP = 8

Почему:

  1. В треугольнике DBCDBC точки MDCM\in DC и NBCN\in BC делят стороны в одинаковой доле от вершины CC:
MCDC=23,CNCB=23.\frac{MC}{DC}=\frac{2}{3},\qquad \frac{CN}{CB}=\frac{2}{3}.

Значит, MNDBMN\parallel DB (теорема Фалеса).

  1. Плоскость α=KMN\alpha=KMN содержит прямую, параллельную DBDB; следовательно, её пересечение с плоскостью ABDABD — прямая, проходящая через KK и параллельная DBDB. Обозначим точку пересечения этой прямой с ABAB как PP.

  2. В треугольнике ABDABD прямые KPDBKP\parallel DB. Тогда

APAB=AKAD=22+3=25.\frac{AP}{AB}=\frac{AK}{AD}=\frac{2}{2+3}=\frac{2}{5}.

При AB=20AB=20:

AP=2025=405=8.AP=20\cdot\frac{2}{5}= \frac{40}{5}=8.

Построение сечения: соединить MM и NN (MNDBMN\parallel DB); через KK провести в плоскости ABDABD прямую KPDBKP\parallel DB, получить P=KPABP=KP\cap AB; затем провести PNPN в плоскости ABCABC и KMKM в плоскости ADCADC. Сечение — четырёхугольник KPNMKPNM.

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