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

2019 AMC 10A Problems/Problem 21

Revision as of 14:40, 22 November 2020 by Awu2014 (talk | contribs) (→‎Solution 4 (similar triangles))
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]

Solution 1

The triangle is placed on the sphere so that its 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 $9$ - $12$ - $15$ allows us easily to determine that the base is $24$ and the height is $9$. The formula $\frac {\text{base} \cdot \text{height}} {2}$ can also be used to find the area of the triangle as $108$, while the semiperimeter is simply $\frac {15 + 15 + 24} {2} = 27$. After plugging into the equation, we thus get $108 = \text{inradius} \cdot 27$, so the inradius is $4$. Now, let the distance between $O$ and the triangle be $x$. Choose a point on the incircle and denote it by $A$. The distance $OA$ is $6$, because it is just the radius of the sphere. The distance from point $A$ to the center of the incircle is $4$, because it is the radius of the incircle. By using the Pythagorean Theorem, we thus find $x = \sqrt{6^2-4^2}=\sqrt{20} = \boxed{\textbf {(D) } 2 \sqrt {5}}$.

Solution 2 (No Inradius)

Drop an altitude on the isosceles triangle. Let the resulting 3-4-5 right triangle $ABC$ have $AB=15$ and $BC=12$. By special triangle, $AC=9$. Let $r$ be the circle's radius. Let the circle's center be $O$ and $D$ be the closest point on $AB$ to $O$.


Then, $OD=r$. Obviously, $ODBC$ is a kite. Thus, $BC=DB=12$, and $AD=15-DB=3$. $AO=AC-r=9-r$. By Pythagoras, $AD^2+OD^2=AO^2$, so $3^2+r^2=(9-r)^2$. The $r^2$ terms cancel out, and $r=4$.


As before, using Pythagoras again, the distance is $\sqrt{6^2-4^2}=\boxed{\textbf{(D)}2\sqrt{5}}$

(Solution by BJHHar)

Solution 3

As in Solution 1, we note that by the Pythagorean Theorem, the height of the triangle is $9$, and that 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, we use the Angle Bisector Theorem. [asy] draw((0,0)--(12,9)--(24,0)--cycle); dot((0,0)); dot((12,9)); dot((24,0)); dot((12,0)); label("$A$",(0,0),SW); label("$B$",(12,9),N); label("$C$",(24,0),SE); label("$D$",(12,-1/2),S); label("$I$",(12,4),SE); draw((12,9)--(12,0), dashed); draw(circle((12,4),4)); draw((0,0)--(216/13,216/39)); dot((12,4)); label("$15$", (6,9/2),NW); label("$12$", (6,-1),S); [/asy] \[\begin{split}&\frac{AB}{BI}=\frac{AD}{DI} \\ \Rightarrow \ &\frac{15}{BI}=\frac{12}{DI} \\ \Rightarrow \ &\frac{BI}{5}=\frac{DI}{4}\end{split}\] Since we know that $BI+DI$ (the height) is equal to $9$, $DI$ (the inradius) is $4$. From here, the problem can be solved in the same way as in Solution 1. The answer is $\boxed{\textbf {(D) } 2 \sqrt{5}}$.

Solution 4 (similar triangles)

First, we label a few points: [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("$r$",(54/5, 28/5), SW); label("$12$", (6,-1),S); label("$I$",(12,4),SE); label("$A$",(0,0),SW); label("$B$",(12,9),N); label("$C$",(24,0),SE); label("$D$",(12,-1/2),S); label("$E$",(48/5, 36/5),NW); [/asy]

We have that $\triangle{BDC}$ is a $3-4-5$ triangle, so, as in Solution 1, $BD = 9$. From this, we know that $\overline{BI}=9-r$. Since $AB$ is tangent to circle $I$, we also know $IEB$ is a right triangle. $\triangle{BIE}$ and $\triangle{BDA}$ share angle $DBA$, so $\triangle{BIE} \sim \triangle{BDA}$ since they have two equal angles. Hence, by this similarity, $\dfrac{9-r}{5}=\dfrac{r}{4}$. Cross-multiplying, we get $36-4r =5r$, which gives $r=4$. We now take another cross section of the sphere, perpendicular to the plane of the triangle.

[asy] draw(circle((6,6),6)); draw((6,6)--(1.75735931,1.75735931)--(6,1.75735931)--cycle); dot((6,6)); dot((1.75735931,1.75735931)); dot((6,1.75735931)); label("$O$", (6,6),N); label("$6$", (3.87867965,3.87867965),NW); label("$4$", (3.87867965,1.75735931),SE); [/asy]

Using the Pythagorean Theorem, we find that the distance from the center to the plane is $\boxed{\textbf {(D) } 2 \sqrt{5}}$.

Community Discussion

https://artofproblemsolving.com/community/c3h1988124

Video Solution

https://www.youtube.com/watch?v=P9-Snytos4o

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=138119"