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This problem trains for: AMC-10, AMC-12.

From a circle with center at O and radius 10 units a portion that spans two-fifths of the circumference is removed. The remaining portion can be furled into a slanted cone with vertex at P. How far from O must P be in order for the cone to have a section that is a right triangle?

The figure below show the circle and the portion that must be removed in order to produce a slanted cone:

figure

For the slanted cone to have a section that is a right angle triangle, it must look like this:

figure

The elements of the section map to the elements of the circle of origin as in:

figure

Therefore, we must have:

equation

If

equation

equation

equation

equation

and using the fact that two-fifths of the circumference have been removed we can find VM:

equation

equation

Solving for x:

equation

equation

equation