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 "1996 AHSME Problems/Problem 27"

(→‎Solution 2)
(→‎Solution 2)
Line 37: Line 37:
 
Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a 2-dimensional plane (the previous z-axis is the new x-axis and the y-axis remains the same).
 
Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a 2-dimensional plane (the previous z-axis is the new x-axis and the y-axis remains the same).
  
−
The spheres now become circles with centers at <math>(1,0)</math> and <math>(21/2,0)</math>.  They have radii <math>9/2</math> and <math>6</math>, respectively.
+
The spheres now become circles with centers at <math>(1,0)</math> and <math>(\frac{21}{2},0)</math>.  They have radii <math>9/2</math> and <math>6</math>, respectively.
 
Let circle <math>A</math> be the circle centered on <math>(1,0)</math> and circle <math>B</math> be the one centered on <math>(21/2,0)</math>.
 
Let circle <math>A</math> be the circle centered on <math>(1,0)</math> and circle <math>B</math> be the one centered on <math>(21/2,0)</math>.
  

Revision as of 21:21, 30 December 2018

Problem

Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls?

$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$

Solution 1

The two equations of the balls are

\[x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \le 36\]

\[x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}\]

Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$, and the second ball goes from $1 \pm 4.5$. The only integer value that $z$ can be is $z=5$.

Plugging that in to both equations, we get:

\[x^2 + y^2 \le \frac{23}{4}\]

\[x^2 + y^2 \le \frac{17}{4}\]

The second inequality implies the first inequality, so the only condition that matters is the second inequality.

From here, we do casework, noting that $|x|, |y| \le 3$:

For $x=\pm 2$, we must have $y=0$. This gives $2$ points.

For $x = \pm 1$, we can have $y\in \{-1, 0, 1\}$. This gives $2\cdot 3 = 6$ points.

For $x = 0$, we can have $y \in \{-2, -1, 0, 1, 2\}$. This gives $5$ points.

Thus, there are $\boxed{13}$ possible points, giving answer $\boxed{D}$.

Solution 2

Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a 2-dimensional plane (the previous z-axis is the new x-axis and the y-axis remains the same).

The spheres now become circles with centers at $(1,0)$ and $(\frac{21}{2},0)$. They have radii $9/2$ and $6$, respectively. Let circle $A$ be the circle centered on $(1,0)$ and circle $B$ be the one centered on $(21/2,0)$.

The point on circle $A$ closest to the center of circle $B$ is $(11/2,0)$. The point on circle B closest to the center of circle $A$ is $(9/2,0)$.

Taking a look back at the 3-dimensional coordinate grid with the spheres, we can see that their intersection appears to be a circle with congruent dome shapes on either end. Because the tops of the domes are at $(0,0,9/2)$ and $(0,0,11/2)$, respectively, the lattice points inside the area of intersection must have z-value $5$ (because $5$ is the only integer between $9/2$ and $11/2$). Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle. The radius of the circle will be the distance from the z-axis.

Now, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles $A$ and $B$ and third point at the first quadrant intersection of the circles.

We know all three side lengths of this triangle: $9/2$ (the radius of circle $A$), $6$ (the radius of circle $B$), and $19/2$ (the distance between the centers of circles $A$ and $B$). Using Heron's formula:

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1996_AHSME_Problems/Problem_27&oldid=99878"