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

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==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>\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 triples <math>9</math> - <math>12</math> - <math>15</math> show the base is <math>24</math> and height is 9. <math>\frac {\text{base} \cdot \text{height}} {2}</math> can be used to find the area of the triangle, <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. Point <math>A</math> to the center of the incircle is length <math>4</math> because it is the inradius of the incircle. By using 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>.
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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>. 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.

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?

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

Diagram

3D [asy] import graph3; import palette; size(200); currentprojection=orthographic(0,4,2);  triple f(pair z) {return expi(z.x,z.y);}  surface s=surface(f,(0,0),(pi,2pi),70,Spline); draw((0,-5/6,sqrt(5)/3)--(2,2/3,sqrt(5)/3)--(-2,2/3,sqrt(5)/3)--cycle); draw(s,mean(palette(s.map(zpart),Grayscale())),nolight); draw((2,2/3,sqrt(5)/3)--(-2,2/3,sqrt(5)/3)); [/asy] Plane through triangle. [asy] draw((0,0)--(12,9)--(24,0)--cycle); draw((12,9)--(12,0), dashed); draw((11.5,0)--(11.5,0.5)--(12,0.5)); draw(circle((12,4),4)); draw((12,4)--(48/5, 36/5)); dot((12,4)); label("$15$", (6,9/2),NW); label("$15$", (18,9/2),NE); label("$24$", (12,-1),S); label("$r$",(54/5, 28/5), SW); [/asy] -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 $\text{area} = \text{inradius} \cdot \text{semiperimeter}$. 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 $24$. The Pythagorean triple $9$ - $12$ - $15$ shows that the base is $24$ and the height is $9$. $\frac {\text{base} \cdot \text{height}} {2}$ can be used to find the area of the triangle as $108$. The semiperimeter is $\frac {15 + 15 + 24} {2} = 27$. After plugging into the equation $108 = \text{inradius} \cdot 27$, we get $\text{inradius} = 4$. Let the distance between $O$ and the triangle be $x$. Choose a point on the incircle and denote it $A$. $\overline{OA}$ is $6$ because it is the radius of the sphere. The distance from point $A$ to the center of the incircle is $4$ because it is the inradius of the incircle. By using the Pythagorean Theorem, you will get that $x$ is $\sqrt{6^2-4^2}=\sqrt{20}\implies\boxed{\textbf {(D) } 2 \sqrt {5}}$.

- ViolinGod(Argonauts16 latex)

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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