Problem:
A sphere with center has radius 6 . A triangle with sides of length 15,15 , and 24 is situated in space so that each of its sides is tangent to the sphere. What is the distance between and the plane determined by the triangle?
Answer Choices:
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Solution:
The triangle is placed on the sphere so that its three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use area inradius semiperimeter. The area of the triangle can be found by drawing an altitude from the vertex between sides with length to the midpoint of the side with length . The Pythagorean triple allows us easily to determine that the base is and the height is . The formula can also be used to find the area of the triangle as , while the semiperimeter is simply . After plugging into the equation, we thus get inradius , so the inradius is . Now, let the distance between and the triangle be . Choose a point on the incircle and denote it by . The distance is , because it is just the radius of the sphere. The distance from point to the center of the incircle is , because it is the radius of the incircle. By using the Pythagorean Theorem, we thus find .
OR
Drop an altitude on the isosceles triangle. Let the resulting right triangle have and . By special triangle, . Let be the circle's radius. Let the circle's center be and be the closest point on to .
Then, . Obviously, is a kite. Thus, , and . . By Pythagoras, , so . The terms cancel out, and .
As before, using Pythagoras again, the distance is .
The problems on this page are the property of the MAA's American Mathematics Competitions