Difference between revisions of "2000 AIME II Problems/Problem 12"
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== Solution == | == Solution == | ||
− | + | Let <math>D</math> be the foot of the perpendicular from <math>O</math> to the plane of <math>ABC</math>. By the [[Pythagorean Theorem]] on triangles <math>\triangle OAD</math>, <math>\triangle OBD</math> and <math>\triangle OCD</math> we get: | |
+ | <math>DA^2=DB^2=DC^2=20^2-OD^2</math> | ||
+ | |||
+ | It follows that <math>DA=DB=DC</math>, so <math>D</math> is the [[circumcenter]] of <math>\triangle ABC</math>. | ||
+ | |||
+ | By [[Heron's Formula]] the area of <math>\triangle ABC</math> is: | ||
+ | |||
+ | <math>K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-15)(21-14)(21-13)} = \sqrt{(21)(8)(7)(6)} = \sqrt{(3\cdot 7)(2^3)(7)(2\cdot 3)} =\sqrt{2^4\cdot 3^2\cdot 7^2} = 2^2\cdot 3\cdot 7 = 84</math> | ||
+ | |||
+ | Now the [[circumradius]] of <math>\triangle ABC</math> is: | ||
+ | |||
+ | <math>R = \frac{abc}{4K} = \frac{(13)(14)(15)}{4(84)} = \frac{65}{8}</math> | ||
+ | |||
+ | Thus by the Pythagorean Theorem again, | ||
+ | |||
+ | <math>OD = \sqrt{20^2-R^2} = \sqrt{20^2-\frac{65^2}{8^2}} = \sqrt{\frac{160^2-65^2}{8^2}} = \frac{\sqrt{(160-65)(160+65)}}{8} = \frac{\sqrt{(95)(225)}}{8} = \frac{15\sqrt{95}}{8}</math>. | ||
+ | |||
+ | So the final answer is <math>15+95+8=\boxed{118}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=II|num-b=11|num-a=13}} | {{AIME box|year=2000|n=II|num-b=11|num-a=13}} |
Revision as of 01:22, 27 November 2007
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 triangle is , where , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .
Solution
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:
Now the circumradius of is:
Thus by the Pythagorean Theorem again,
.
So the final answer is
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 |