Difference between revisions of "2007 AIME I Problems/Problem 15"
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− | Use the [[quadratic formula]], though we only need the root of the [[discriminant]]. This is <math>\sqrt{(7 \cdot 66)^2 - 4 \cdot 10 \cdot (66^2 + 90)} = \sqrt{49 \cdot 66^2 - 40 \cdot 66^2 - 4 \cdot 9 \cdot 100}</math><math> = \sqrt{9 \cdot 4 \cdot 33^2 - 9 \cdot 4 \cdot 100} = 6\sqrt{33^2 - 100}</math>. The answer is <math>989</math>. | + | Use the [[quadratic formula]], though we only need the root of the [[discriminant]]. This is <math>\sqrt{(7 \cdot 66)^2 - 4 \cdot 10 \cdot (66^2 + 90)} = \sqrt{49 \cdot 66^2 - 40 \cdot 66^2 - 4 \cdot 9 \cdot 100}</math><math> = \sqrt{9 \cdot 4 \cdot 33^2 - 9 \cdot 4 \cdot 100} = 6\sqrt{33^2 - 100}</math>. The answer is <math>\boxed{989}</math>. |
== See also == | == See also == |
Revision as of 17:39, 18 February 2019
Problem
Let be an equilateral triangle, and let
and
be points on sides
and
, respectively, with
and
. Point
lies on side
such that angle
. The area of triangle
is
. The two possible values of the length of side
are
, where
and
are rational, and
is an integer not divisible by the square of a prime. Find
.
Solution
Denote the length of a side of the triangle , and of
as
. The area of the entire equilateral triangle is
. Add up the areas of the triangles using the
formula (notice that for the three outside triangles,
):
. This simplifies to
. Some terms will cancel out, leaving
.
is an external angle to
, from which we find that
, so
. Similarly, we find that
. Thus,
. Setting up a ratio of sides, we get that
. Using the previous relationship between
and
, we can solve for
.
Use the quadratic formula, though we only need the root of the discriminant. This is . The answer is
.
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.