Difference between revisions of "2015 AIME II Problems/Problem 15"
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The area | The area | ||
<cmath>[ACE] = [AQC]+[CQE]– [AQE].</cmath> | <cmath>[ACE] = [AQC]+[CQE]– [AQE].</cmath> | ||
− | Hence <cmath>[ACE] =\frac{AQ^2}{2} (\sin 2\alpha + \sin 2\beta - \sin(2\alpha + 2\beta)),</cmath> | + | Hence <cmath>[ACE] =\frac{AQ^2}{2} \left(\sin 2\alpha + \sin 2\beta - \sin(2\alpha + 2\beta)\right),</cmath> |
<cmath>[ACE] = | <cmath>[ACE] = | ||
− | 8( \frac{16}{65}+\frac{4}{5} - \frac{4}{5}\cdot \frac{63}{65} - \frac{3}{5}\cdot \frac{16}{65} = \frac{64}{65}\implies \boxed {129}.</cmath> | + | 8\left( \frac{16}{65}+\frac{4}{5} - \frac{4}{5}\cdot \frac{63}{65} - \frac{3}{5}\cdot \frac{16}{65}\right) = \frac{64}{65}\implies \boxed {129}.</cmath> |
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2015|n=II|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:22, 7 July 2022
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
is your friend for a quick solve. If you know about homotheties, go ahead, but you'll still need to do quite a bit of computation. If you're completely lost and you have a lot of time left in your mocking of this AIME, go ahead and use analytic geometry.
Solution 1 (guys trig is fast)
Let be the intersection of
and the common internal tangent of
and
We claim that
is the circumcenter of right
Indeed, we have
and
by equal tangents to circles, and since
is the midpoint of
implying that
Now draw
where
is the center of circle
Quadrilateral
is cyclic, and by Pythagorean Theorem
so by Ptolemy on
we have
Do the same thing on cyclic quadrilateral
(where
is the center of circle
and get
Let By Law of Sines,
Note that
from inscribed angles, so
after angle addition identity.
Similarly, and by Law of Sines
Note that
from inscribed angles, so
after angle addition identity.
Setting the two areas equal, we get
after Pythagorean Identity. Now plug back in and the common area is
Solution 2
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 3
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
Alternate Path to x
Call the intersection of lines and
.You can use similar triangles to find that the distance from
to
is four times the distance from
to
. Then draw a perpendicular from
to
and call the point
.
and
, so by the Pythagorean Theorem,
. You can now use similar triangles to find that
and continue on like in solution 2.
Solution 4
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
.
Solution 5 (HARD computation)
Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because and
,
is right angle.
First, , so
. And,
Then,
Since
,
,
, we have
Since
is four times in scale to
, their area ratio is 16. Divide the two equations for the two areas, we have
With this angle found, everything else just follows.
Solution 6 (Simple computation)
Let be the intersection of
and
. Since the radii of the two circles are 1:4, so we have
, and the distance from
to line
and the distance from
to line
are in a ratio of 4:1, so
. We can easily calculate the length of
to be 4, so
. Let
be the foot of perpendicular line from
to
, we can know that
, so
,
,
, and
. Since
, so
, and
.
, so the distance from
to line
is
. so the area is
The final answer is
.
--- by Dan Li
Solution 7
Consider the common tangent from to both circles. Let this intersect
at point
. From equal tangents, we have
, which implies that
.
Let the center of be
, and the center of
be
. Angle chasing, we find that
with a ratio of
. Hence
.
We can easily deduce that by dropping an altitude from
to
. Let
. By some simple angle chasing, we obtain that
and similarly
.
Using LoC, we get that and
. From Pythagorean theorem, we have
In other words,
.
Using the area condition, we have:
Now, for brevity, let and
.
From Law of Sines on , we have
It remains to find the area of . This is just
for an answer of
This solution was brought to you by Leonard_my_dude.
Solution 8 (Synthetic-Trigonometry)
Add in the line as the internal tangent between the two circles. Let
be the midpoint of
; It is well-known that
is on
and because
is the radical axis of the two circles,
. Therefore because
is the circumcenter of
,
. Let
be the center of circle
and likewise let
be the center of circle
. It is well known that by homothety
and
are collinear. It is well-known that
, and likewise
. By homothety,
, therefore since the two triangles mentioned in the problem, the length of the altitude from
to
is four times the length of the altitude from
to
. Using the Pythagorean Theorem,
. By angle-chasing,
is cyclic, and likewise
is cyclic. Use the Pythagorean Theorem for
to get
. Then by Ptolemy's Theorem
. Now to compute the area, using what we know about the length of the altitude from
to
is four times the length of the altitude from
to
, letting
be the length of the altitude from
to
,
. From the Law of Sines,
. Then use the Pythagorean Theorem twice and add up the lengths to get
. Use the formula
to get
as the answer.
~First
Solution 9 (Visual)
Let and
be the centers of circles
and
, respectively.
Let be midpoint
Upper diagram shows that
and
Therefore
Let Lower diagram shows that
(perpendicular sides)
and (the same intersept
The area
Hence
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
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