Difference between revisions of "2006 AIME II Problems/Problem 9"
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The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</math>. | The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</math>. | ||
− | From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{ | + | From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{4}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{3}}</math>. |
The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} </math> <math>\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}</math> <math>= \frac{76 - 12\sqrt{5}}{4}</math> <math>= 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow \boxed{027}</math>. | The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} </math> <math>\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}</math> <math>= \frac{76 - 12\sqrt{5}}{4}</math> <math>= 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow \boxed{027}</math>. |
Latest revision as of 10:45, 26 August 2015
Problem
Circles and
have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line
is a common internal tangent to
and
and has a positive slope, and line
is a common internal tangent to
and
and has a negative slope. Given that lines
and
intersect at
and that
where
and
are positive integers and
is not divisible by the square of any prime, find
Solution
Call the centers , the points of tangency
(with
on
and
on
, and
on
), and the intersection of each common internal tangent to the X-axis
.
since both triangles have a right angle and have vertical angles, and the same goes for
. By proportionality, we find that
; solving
by the Pythagorean theorem yields
. On
, we can do the same thing to get
and
.
The vertical altitude of each of and
can each by found by the formula
(as both products equal twice of the area of the triangle). Thus, the respective heights are
and
. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem:
, and by 30-60-90:
.
From this information, the slope of each tangent can be uncovered. The slope of . The slope of
.
The equation of can be found by substituting the point
into
, so
. The equation of
, found by substituting point
, is
. Putting these two equations together results in the desired
. Thus,
.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.