Difference between revisions of "2000 AIME II Problems/Problem 12"
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== Solution 2 (Vectors) == | == Solution 2 (Vectors) == | ||
− | We know the radii to <math>A</math>,<math>B</math>, and <math>C</math> form a triangular pyramid <math>OABC</math>. We know the lengths of the edges <math>OA = OB = OC = 20</math>. First we can break up <math>ABC</math> into its two component right triangles <math>5-12-13</math> and <math>9-12-15</math>. Let the <math>y</math> axis be perpendicular to the base and <math>x</math> axis run along <math>BC</math>, and <math>z</math> occupy the other dimension, with the origin as <math>C</math>. We look at vectors <math>OA</math> and <math>OC</math>. Since <math>OAC</math> is isoceles we know the vertex is equidistant from <math>A</math> and <math>C</math>, hence it is <math>7</math> units along the <math>x</math> axis. Hence for vector <math>OC</math>, in form <math><x,y,z></math> it is <math><7, h, l</math> where <math>h</math> is the height (answer) and <math>l</math> is the component of the vertex along the <math>z</math> axis. Now on vector <math>OA</math>, since <math>A</math> is <math>9</math> along <math>x</math>, and it is <math>12</math> along <math>z</math> axis, it is <math><-2, h, 12- l></math>. We know both vector magnitudes are <math>20</math>. Solving for <math>h</math> yields <math>\frac{15\sqrt{95} }{8}</math>, so Answer = <math>\boxed{118}</math>. | + | We know the radii to <math>A</math>,<math>B</math>, and <math>C</math> form a triangular pyramid <math>OABC</math>. We know the lengths of the edges <math>OA = OB = OC = 20</math>. First we can break up <math>ABC</math> into its two component right triangles <math>5-12-13</math> and <math>9-12-15</math>. Let the <math>y</math> axis be perpendicular to the base and <math>x</math> axis run along <math>BC</math>, and <math>z</math> occupy the other dimension, with the origin as <math>C</math>. We look at vectors <math>OA</math> and <math>OC</math>. Since <math>OAC</math> is isoceles we know the vertex is equidistant from <math>A</math> and <math>C</math>, hence it is <math>7</math> units along the <math>x</math> axis. Hence for vector <math>OC</math>, in form <math><x,y,z></math> it is <math><7, h, l></math> where <math>h</math> is the height (answer) and <math>l</math> is the component of the vertex along the <math>z</math> axis. Now on vector <math>OA</math>, since <math>A</math> is <math>9</math> along <math>x</math>, and it is <math>12</math> along <math>z</math> axis, it is <math><-2, h, 12- l></math>. We know both vector magnitudes are <math>20</math>. Solving for <math>h</math> yields <math>\frac{15\sqrt{95} }{8}</math>, so Answer = <math>\boxed{118}</math>. |
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | this is highly trivial for an AIME #12 |
Latest revision as of 13:14, 22 August 2020
Problem
The points ,
and
lie on the surface of a sphere with center
and radius
. It is given that
,
,
, and that the distance from
to
is
, where
,
, and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
.
Solution 1
Let be the foot of the perpendicular from
to the plane of
. By the Pythagorean Theorem on triangles
,
and
we get:
It follows that , so
is the circumcenter of
.
By Heron's Formula the area of is (alternatively, a
triangle may be split into
and
right triangles):
From , we know that the circumradius of
is:
Thus by the Pythagorean Theorem again,
So the final answer is .
Solution 2 (Vectors)
We know the radii to ,
, and
form a triangular pyramid
. We know the lengths of the edges
. First we can break up
into its two component right triangles
and
. Let the
axis be perpendicular to the base and
axis run along
, and
occupy the other dimension, with the origin as
. We look at vectors
and
. Since
is isoceles we know the vertex is equidistant from
and
, hence it is
units along the
axis. Hence for vector
, in form
it is
where
is the height (answer) and
is the component of the vertex along the
axis. Now on vector
, since
is
along
, and it is
along
axis, it is
. We know both vector magnitudes are
. Solving for
yields
, so Answer =
.
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
this is highly trivial for an AIME #12