Difference between revisions of "2019 AMC 10A Problems/Problem 21"
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| − | 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 <math> area = inradius * semiperimeter </math>. The area of the triangle can be found by drawing an altitude from the vertex between sides with length 15 to the midpoint of the side with length 24. Pythagorean triples 9 - 12 - 15 show the base is 24 and height is 9. <math>\frac {base * height} {2}</math> can be used to find the area of the triangle, 108. The semiperimeter is <math>\frac {15 + 15 + 24} {2} = 27</math> After plugging into the equation <math>108 = inradius * 27</math>, the inradius = 4. Let the distance between O and the triangle be x. Choose a point on the incircle A. OA is 6 because it is the radius of the sphere. Point A to the center of the incircle measures 4 because it is the inradius of the incircle. By using Pythagorean Theorem, you will get that the distance between x and O is <math>\textbf {(D)} 2 \sqrt {5}</math> | + | 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 <math> area = inradius * semiperimeter </math>. The area of the triangle can be found by drawing an altitude from the vertex between sides with length 15 to the midpoint of the side with length 24. Pythagorean triples 9 - 12 - 15 show the base is 24 and height is 9. <math>\frac {base * height} {2}</math> can be used to find the area of the triangle, 108. The semiperimeter is <math>\frac {15 + 15 + 24} {2} = 27</math> After plugging into the equation <math>108 = inradius * 27</math>, the inradius = 4. Let the distance between O and the triangle be x. Choose a point on the incircle A. OA is 6 because it is the radius of the sphere. Point A to the center of the incircle measures 4 because it is the inradius of the incircle. By using Pythagorean Theorem, you will get that the distance between x and O is <math>\textbf {(D)} 2 \sqrt {5}</math> |
- ViolinGod | - ViolinGod | ||
Revision as of 16:35, 9 February 2019
- The following problem is from both the 2019 AMC 10A #21 and 2019 AMC 12A #18, so both problems redirect to this page.
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?
Solution
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 15 to the midpoint of the side with length 24. Pythagorean triples 9 - 12 - 15 show the base is 24 and height is 9. 
can be used to find the area of the triangle, 108. The semiperimeter is 
After plugging into the equation 
, the inradius = 4. Let the distance between O and the triangle be x. Choose a point on the incircle A. OA is 6 because it is the radius of the sphere. Point A to the center of the incircle measures 4 because it is the inradius of the incircle. By using Pythagorean Theorem, you will get that the distance between x and O is 
- ViolinGod
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
