Difference between revisions of "2012 AIME II Problems/Problem 15"
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'''-Solution by thecmd999''' | '''-Solution by thecmd999''' | ||
+ | ==Solution 3== | ||
+ | <asy> | ||
+ | pair E; | ||
+ | real z=3^.5*14/3; | ||
+ | E=(z,0); | ||
+ | </asy> | ||
== 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 13:24, 26 March 2016
Contents
[hide]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
pair E; real z=3^.5*14/3; E=(z,0); (Error making remote request. Unknown error_msg)
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 |
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