Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 10A Problems/Problem 21"

(→‎Solution)
(→‎Solution)
Line 14: Line 14:
  
 
==Solution==
 
==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 <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>\fbox \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 17:53, 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 $O$ 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 $O$ and the plane determined by the triangle?

$\textbf{(A) }2\sqrt{3}\qquad \textbf{(B) }4\qquad \textbf{(C) }3\sqrt{2}\qquad \textbf{(D) }2\sqrt{5}\qquad \textbf{(E) }5\qquad$

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 $area = inradius * semiperimeter$. 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. $\frac {base * height} {2}$ can be used to find the area of the triangle, 108. The semiperimeter is $\frac {15 + 15 + 24} {2} = 27$ After plugging into the equation $108 = inradius * 27$, 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 $\textbf {(D)} 2 \sqrt {5}$.

- 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_21&oldid=101540"