Difference between revisions of "1996 AHSME Problems/Problem 27"
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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>(\frac{21}{2},0)</math>. They have radii <math>9 | + | The spheres now become circles with centers at <math>(1,0)</math> and <math>(\frac{21}{2},0)</math>. They have radii <math>\frac{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 | + | 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>(\frac{21}{2},0)</math>. |
− | The point on circle <math>A</math> closest to the center of circle <math>B</math> is <math>(11 | + | The point on circle <math>A</math> closest to the center of circle <math>B</math> is <math>(\frac{11}{2},0)</math>. The point on circle B closest to the center of circle <math>A</math> is <math>(\frac{11}{2},0)</math>. |
− | 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 <math>(0,0,9 | + | 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 <math>(0,0,\frac{9}{2})</math> and <math>(0,0,\frac{11}{2})</math>, respectively, the lattice points inside the area of intersection must have z-value <math>5</math> (because <math>5</math> is the only integer between <math>\frac{9}{2}</math> and <math>\frac{11}{2}</math>). 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 <math>A</math> and <math>B</math> and third point at the first quadrant intersection of the circles. | 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 <math>A</math> and <math>B</math> and third point at the first quadrant intersection of the circles. | ||
− | We know all three side lengths of this triangle: <math>9 | + | We know all three side lengths of this triangle: <math>\frac{9}{2}</math> (the radius of circle <math>A</math>), <math>6</math> (the radius of circle <math>B</math>), and <math>\frac{19}{2}</math> (the distance between the centers of circles <math>A</math> and <math>B</math>). We can now find the area of the triangle using Heron's formula: |
+ | |||
+ | <cmath>s=\frac{\frac{9}{2}+6+\frac{19}{2}}{2}=20</cmath> | ||
+ | |||
+ | <cmath>A=\sqrt{(20)(20-\frac{9}{2})(20-6)(20-\frac{19}{2})}</cmath> | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=26|num-a=28}} | {{AHSME box|year=1996|num-b=26|num-a=28}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:25, 30 December 2018
Contents
Problem
Consider two solid spherical balls, one centered at with radius
, and the other centered at
with radius
. How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
Solution 1
The two equations of the balls are
Note that along the axis, the first ball goes from
, and the second ball goes from
. The only integer value that
can be is
.
Plugging that in to both equations, we get:
The second inequality implies the first inequality, so the only condition that matters is the second inequality.
From here, we do casework, noting that :
For , we must have
. This gives
points.
For , we can have
. This gives
points.
For , we can have
. This gives
points.
Thus, there are possible points, giving answer
.
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 and
. They have radii
and
, respectively.
Let circle
be the circle centered on
and circle
be the one centered on
.
The point on circle closest to the center of circle
is
. The point on circle B closest to the center of circle
is
.
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 and
, respectively, the lattice points inside the area of intersection must have z-value
(because
is the only integer between
and
). 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 and
and third point at the first quadrant intersection of the circles.
We know all three side lengths of this triangle: (the radius of circle
),
(the radius of circle
), and
(the distance between the centers of circles
and
). We can now find the area of the triangle 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.