Difference between revisions of "2012 AIME II Problems/Problem 15"
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However, the problem asks for <math>AF^2</math>, so <math>AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}</math>. | However, the problem asks for <math>AF^2</math>, so <math>AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}</math>. | ||
− | + | '''-Solution by TheBoomBox77''' | |
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:04, 27 March 2016
Problem 15
Triangle is inscribed in circle
with
,
, and
. The bisector of angle
meets side
at
and circle
at a second point
. Let
be the circle with diameter
. Circles
and
meet at
and a second point
. Then
, where
and
are relatively prime positive integers. Find
.
Solution 1
Use the angle bisector theorem to find ,
, and use the Stewart's Theorem to find
. Use Power of the Point to find
, and so
. Use law of cosines to find
, hence
as well, and
is equilateral, so
.
I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:
(1)
Adding these two and simplifying we get:
(2). Ah, but
(since
lies on
), and we can find
using the law of cosines:
, and plugging in
we get
.
Also, , and
(since
is on the circle
with diameter
), so
.
Plugging in all our values into equation (2), we get:
, or
.
Finally, we plug this into equation (1), yielding:
. Thus,
or
The answer is
.
Solution 2
Let ,
,
for convenience. We claim that
is a symmedian. Indeed, let
be the midpoint of segment
. Since
, it follows that
and consequently
. Therefore,
. Now let
. Since
is a diameter,
lies on the perpendicular bisector of
; hence
,
,
are collinear. From
, it immediately follows that quadrilateral
is cyclic. Therefore,
, implying that
is a symmedian, as claimed.
The rest is standard; here's a quick way to finish. From above, quadrilateral is harmonic, so
. In conjunction with
, it follows that
. (Notice that this holds for all triangles
.) To finish, substitute
,
,
to obtain
as before.
-Solution by thecmd999
Solution 3
First of all, use the Angle Bisector Theorem to find that
and
, and use Stewart's Theorem to find that
. Then use Power of a Point to find that
. Then use the circumradius of a triangle formula to find that the length of the circumradius of
is
.
Since DE is the diameter of circle ,
is
. Extending
to intersect circle
at
, we find that
is the diameter of the circumcircle of
(since
is
). Therefore,
.
Let ,
, and
. Then, by the Pythagorean Theorem,
and
Subtracting the first equation from the second, the term cancels out and we obtain:
By Power of a Point, , so
Since ,
.
Because and
intercept the same arc in circle
and the same goes for
and
,
by AA Similarity. Since side lengths in similar triangles are proportional,
However, the problem asks for , so
.
-Solution by TheBoomBox77
See Also
2012 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.