Difference between revisions of "2015 AIME II Problems/Problem 15"
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− | Hence, we need to find the slope <math>m</math> of line <math>\ell</math> such that the perpendicular distance <math>n</math> from <math>B</math> to <math>AD</math> is four times the perpendicular distance <math>p</math> from <math>C</math> to <math>AE</math>. This will mean that the product of the bases and heights of triangles <math>ACE</math> and <math>DBA</math> will be equal, which in turn means that their areas will be equal. Let the line <math>\ell</math> have the equation <math>y = -mx \implies mx + y = 0</math> | + | Hence, we need to find the slope <math>m</math> of line <math>\ell</math> such that the perpendicular distance <math>n</math> from <math>B</math> to <math>AD</math> is four times the perpendicular distance <math>p</math> from <math>C</math> to <math>AE</math>. This will mean that the product of the bases and heights of triangles <math>ACE</math> and <math>DBA</math> will be equal, which in turn means that their areas will be equal. Let the line <math>\ell</math> have the equation <math>y = -mx \implies mx + y = 0</math>, and let <math>m</math> be a positive real number so that the negative slope of <math>\ell</math> is preserved. Then, the coordinates of <math>B</math> are <math>(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)</math>, and the coordinates of <math>C</math> are <math>(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)</math>. Using the point-to-line distance formula and the fact that <math>n = 4p</math>, we have <cmath>\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}</cmath> <cmath>\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.</cmath> Since <math>m</math> takes on a positive value, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have <cmath>\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.</cmath> Thus, the equation of <math>\ell</math> is <math>y = -\frac{3}{2}x</math>. |
Revision as of 19:06, 31 March 2015
Contents
Problem
Circles and
have radii
and
, respectively, and are externally tangent at point
. Point
is on
and point
is on
so that line
is a common external tangent of the two circles. A line
through
intersects
again at
and intersects
again at
. Points
and
lie on the same side of
, and the areas of
and
are equal. This common area is
, where
and
are relatively prime positive integers. Find
.
Hint
This is a #15 on an AIME, so it must be difficult. Indeed, there are two possible approaches (both of them very computational): coordinate geometry, or regular Euclidean geometry combined with a bit of trigonometry.
Solution 1
Call and
the centers of circles
and
, respectively, and call
and
the feet of the altitudes from
to
and
to
, respectively. Extend
and
to meet at point
. Using the fact that
and setting
, we have that
. We can do some more length chasing using triangles similar to
to get that
,
, and
. Now, consider the circles
and
on the coordinate plane, where
is the origin. If the line
through
intersects
at
and
at
, then
. To verify this, notice that
from the fact that both triangles are isosceles with
, which are corresponding angles. Since
, we can conclude that
.
Hence, we need to find the slope of line
such that the perpendicular distance
from
to
is four times the perpendicular distance
from
to
. This will mean that the product of the bases and heights of triangles
and
will be equal, which in turn means that their areas will be equal. Let the line
have the equation
, and let
be a positive real number so that the negative slope of
is preserved. Then, the coordinates of
are
, and the coordinates of
are
. Using the point-to-line distance formula and the fact that
, we have
Since
takes on a positive value, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have
Thus, the equation of
is
.
Then we can find the coordinates of by finding the point
other than
where the circle
intersects
.
can be represented with the equation
, and substituting
into this equation yields
Discarding
, the
-coordinate of
is
. The distance from
to
is then
The perpendicular distance from
to
or the height of
is
Finally, the common area is
, and
.
Solution 2
By homothety, we deduce that . (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of
and
to
.) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from
to
is four times that from
to
. Let the distance from
be
and the distance from
be
.
Let and
be the centers of their respective circles. Then dropping a perpendicular from
to
creates a
right triangle, from which
and, if
, that
. Then
, and the Law of Cosines on triangles
and
gives
and
Now, using the Pythagorean Theorem to express the length of the projection of onto line
gives
Squaring and simplifying gives
and squaring and solving gives
By the Law of Sines on triangle , we have
But we know
, and so a small computation gives
The Pythagorean Theorem now gives
and so the common area is
The answer is
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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