2019 AMC 10A Problems/Problem 21
- The following problem is from both the 2019 AMC 10A #21 and 2019 AMC 12A #18, so both problems redirect to this page.
Contents
Problem
A sphere with center has radius
. A triangle with sides of length
and
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?
Diagram
3D
Plane through triangle.
Solution 1
The triangle is placed on the sphere so that 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 . 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
-
-
shows that the base is
and the height is
.
can be used to find the area of the triangle as
. The semiperimeter is
. After plugging into the equation
, we get
. Let the distance between
and the triangle be
. Choose a point on the incircle and denote it
.
is
because it is the radius of the sphere. The distance from point
to the center of the incircle is
because it is the inradius of the incircle. By using the Pythagorean Theorem, you will get that
is
.
Solution 2 (borderline guessing)
Test all the answer choices by plugging them into the expression to find the inradius of the triangle. Seeing that only
gives an integer inradius, we pick
.
Solution 3
According to the Pythagorean Theorem, the height of the triangle is .
The three sides of the triangle are tangent to the sphere, so the circle in the cross-section of the sphere is the incenter of the triangle.
Recall that the inradius is the intersection of the angle bisectors. To find the inradius of the incircle, use the angle bisector theorem.
Since we know that
(the height) is equal to
,
(the inradius) is
.
From here, the problem can be solved in the same way as in Solution 1. The answer is
.
Solution 4 (Similar Triangles)
First, we label a few points:
We know is a
triangle meaning that
.
From this, we know that
.
Since AB is tangent to circle I, we know
is a right triangle.
and
share angle
so
by AA similarity.
As a result, we know
. Cross-multiplying, we get
which gives us
.
We take another cross section of the sphere perpendicular to the plane of the triangle.
Using Pythagorean Theorem, we find that the distance from the center to the plane is .
-woofle628
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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