Very interesting, a pyramid and sphere inscribed

[JuanR]

A pyramid ABCDS is given (the base is convex quadrilateral). A sphere is inscribed in this pyramid and it is tangent to side ABCD at point P.
Prove that
\angle APB + \angle CPD = 180^{o}

 

[JuanR]

* 180^{o} means 180 degree

 

[Architeuthis Dux]

This is a fascinating geometric problem that involves the relationship between a pyramid and a sphere. The fact that the sphere is inscribed in the pyramid and tangent to one of its sides adds an extra layer of complexity to the problem.

To prove that \angle APB + \angle CPD = 180^{o}, we can use the fact that the angle between a tangent line and a radius of a circle is always 90 degrees. In this case, the line segment AP is a tangent to the sphere at point P, and the line segment PB is a radius of the sphere. Similarly, the line segment CP is a tangent to the sphere at point P, and the line segment PD is a radius of the sphere.

Therefore, we can form right triangles APB and CPD, with the right angles at points P. By the Pythagorean theorem, we know that the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. In this case, the legs AP and PB have lengths equal to the radius of the inscribed sphere, and the hypotenuse AB has length equal to the distance between the tangency points of the sphere on sides AB and CD. Similarly, the legs CP and PD have lengths equal to the radius of the inscribed sphere, and the hypotenuse CD has length equal to the distance between the tangency points of the sphere on sides AB and CD.

Therefore, we can write the following equations:

AP^2 + PB^2 = AB^2
CP^2 + PD^2 = CD^2

Adding these two equations together, we get:

AP^2 + PB^2 + CP^2 + PD^2 = AB^2 + CD^2

But we know that AB = CD, since they are both sides of the same base of the pyramid. Therefore, we can simplify the equation to:

AP^2 + PB^2 + CP^2 + PD^2 = 2AB^2

Finally, we can use the fact that the sum of the squares of the sides of a convex quadrilateral is equal to the sum of the squares of its diagonals. In this case, the diagonals of the quadrilateral ABCD are AC and BD. Therefore, we can write the following equation:

AB^2 + CD^2 = AC^2 + BD^2

Substituting this into our previous equation, we get:

AP