Difference between revisions of "2015 AIME II Problems/Problem 15"
m (→Solution 1) |
m (→Hint) |
||
Line 19: | Line 19: | ||
==Hint== | ==Hint== | ||
− | This is a #15 on an AIME, so it must be difficult. Indeed, there are | + | This is a #15 on an AIME, so it must be difficult. Indeed, although there are several possible approaches, all of them are very computational. |
==Solution 1== | ==Solution 1== |
Revision as of 11:54, 10 August 2017
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, although there are several possible approaches, all of them are very computational.
Solution 1
Call and
the centers of circles
and
, respectively, and extend
and
to meet at point
. Call
and
the feet of the altitudes from
to
and
to
, respectively. 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. Setting
, the coordinates of
are
, and the coordinates of
are
. Using the point-to-line distance formula and the condition
, we have
If
, then clearly
and
would not lie on the same side of
. Thus since
, 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
as solutions. 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
Solution 3
goes through
, the point of tangency of both circles. So
intercepts equal arcs in circle
and
: homothety. Hence,
. We will use such similarity later.
The diagonal distance between the centers of the circles is . The difference in heights is
. So
.
The triangle connecting the centers with a side parallel to is a
right triangle. Since
, the height of
is
. Drop an altitude from
to
and call it
:
and
. Since right
,
is a right triangle also;
form a geometric progression
.
Extend through
to a point
on the other side of
. By homothety,
. By angle chasing
through right triangle
, we deduce that
is a right angle. Since
is cyclic,
is also right. So
is a diameter of
. Because of this,
, the tangent line.
is right and
.
so
and
.
Since , the common area is
.
because the triangles are similar with a ratio of
. So we only need to find
now.
Extend through
to intersect the tangent at
. Because
, the altitude from
to
is
times the height from
to
. So
and
. We look at right triangle
.
and
.
is a
right triangle. Hypotenuse
intersects
at a point, we call it
.
. So
.
By Power of a Point, .
. So
. The height from
to
is
.
Thus, . The area of the whole cyclic quadrilateral is
. Lastly, the common area is
the area of the quadrilateral, or
. So
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.