# Download e-book for kindle: Problems in plane and solid geometry. Solid geometry by Prasolov V.V., Sharygin I.F.

By Prasolov V.V., Sharygin I.F.

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Thus, P = (R cos ϕ cos ψ, R cos ϕ sin ψ, R sin ϕ). 23. 22. If the latitude and the longitude of point P are equal to ϕ, then P = (R cos2 ϕ, R cos ϕ sin ϕ, R sin ϕ). The coordinates of the projection of this point to the equatorial plane are x = R cos2 ϕ and y = R cos ϕ sin ϕ. , the set to be found is the circle of radius 12 R centered at ( 12 R, 0). 24. First, let us consider the truncated cone whose lateral surface is tangent to the ball of radius R and center O and let the tangent points divide the generators of the cone in halves.

Then z = R sin ϕ and OP ′ = R cos ϕ. Hence, x = OP ′ cos ψ = R cos ϕ cos ψ; y = R cos ϕ sin ψ. Thus, P = (R cos ϕ cos ψ, R cos ϕ sin ψ, R sin ϕ). 23. 22. If the latitude and the longitude of point P are equal to ϕ, then P = (R cos2 ϕ, R cos ϕ sin ϕ, R sin ϕ). The coordinates of the projection of this point to the equatorial plane are x = R cos2 ϕ and y = R cos ϕ sin ϕ. , the set to be found is the circle of radius 12 R centered at ( 12 R, 0). 24. First, let us consider the truncated cone whose lateral surface is tangent to the ball of radius R and center O and let the tangent points divide the generators of the cone in halves.

60. A plane intersects edges AB, BC, CD and DA of tetrahedron ABCD at points K, L, M and N , respectively; P is an arbitrary point in space. Lines P K, P L, P M and P N intersect the circles circumscribed about triangles P AB, P BC, P CD and P DA for the second time at points K1 , L1 , M1 and N1 , respectively. Prove that points P , K1 , L1 , M1 and N1 lie on one sphere. 1. First, let us prove that the length of the common tangent to the two tangent √ circles of radii R and r is equal to 2 Rr. To this end, let us consider a right triangle the endpoints of whose hypothenuse are the centers of circles and one of the legs is parallel to the common tangents.