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>\text{area} = \text{inradius} \cdot \text{semiperimeter}</math>. The area of the triangle can be found by drawing an altitude from the vertex between sides with length <math>15</math> to the midpoint of the side with length <math>24</math>. Pythagorean | + | 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>\text{area} = \text{inradius} \cdot \text{semiperimeter}</math>. The area of the triangle can be found by drawing an altitude from the vertex between sides with length <math>15</math> to the midpoint of the side with length <math>24</math>. The Pythagorean triple <math>9</math> - <math>12</math> - <math>15</math> shows that the base is <math>24</math> and the height is <math>9</math>. <math>\frac {\text{base} \cdot \text{height}} {2}</math> can be used to find the area of the triangle as <math>108</math>. The semiperimeter is <math>\frac {15 + 15 + 24} {2} = 27</math>. After plugging into the equation <math>108 = \text{inradius} \cdot 27</math>, we get <math>\text{inradius} = 4</math>. Let the distance between <math>O</math> and the triangle be <math>x</math>. Choose a point on the incircle and denote it <math>A</math>. <math>\overline{OA}</math> is <math>6</math> because it is the radius of the sphere. The distance from point <math>A</math> to the center of the incircle is <math>4</math> because it is the inradius of the incircle. By using the Pythagorean Theorem, you will get that <math>x</math> is <math>\sqrt{6^2-4^2}=\sqrt{20}\implies\boxed{\textbf {(D) } 2 \sqrt {5}}</math>. |
- ViolinGod(Argonauts16 latex) | - ViolinGod(Argonauts16 latex) |
Revision as of 19:00, 10 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.
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. -programjames1
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 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 .
- ViolinGod(Argonauts16 latex)
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.