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Mathematics competition in grade three. Get extra points for correct answers.
Solution:

In scheme A, as shown in the figure, O is connected with AO as OE⊥AD in E, and the radius of circle O is R.

Obviously, the triangle AOE is an isosceles right triangle, and AO = 16+R, AE =16-R.

So 16+r = √ 2 * (16-r)

So r = 16 (3-2 √ 2)

So the circumference o of the circle is 32 (3-2 √ 2) π.

And the length of the arc BD = 16 * 2 * π/4 = 8 π.

Therefore, the length of the arc BD is not equal to the circumference of the circle O, and the bottom surface and the side surface cannot be spliced.

So plan a is not feasible.

Scheme b is feasible.

In scheme B, as shown in the figure, O is connected as OE⊥AD in E. Let the radius of circle o be r and the radius of MN arc be X.

In order to make a cone, the length MN of the arc must be equal to the circumference o of the circle.

So 2 * x * π/4 = 2 * π * R.

So X=4R

Obviously, the triangle AOE is an isosceles right triangle, and AO = 5R, AE =16-R.

So 5r = √ 2 * (16-r)

So the radius of the base circle is: r = 16 (5 √ 2-2)/23.

The length of the generatrix of the cone is 16-R = (400-80 √ 2)/23.

According to the above data, we can splice a cone with qualified surfaces.

References:

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