Difference between revisions of "2012 AIME II Problems/Problem 15"
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== Solution 1== | == Solution 1== | ||
− | Use the angle bisector theorem to find <math>CD=\frac{21}{8}</math>, <math>BD=\frac{35}{8}</math>, and use | + | Use the angle bisector theorem to find <math>CD=\frac{21}{8}</math>, <math>BD=\frac{35}{8}</math>, and use Stewart's Theorem to find <math>AD=\frac{15}{8}</math>. Use Power of the Point to find <math>DE=\frac{49}{8}</math>, and so <math>AE=8</math>. Use law of cosines to find <math>\angle CAD = \frac{\pi} {3}</math>, hence <math>\angle BAD = \frac{\pi}{3}</math> as well, and <math>\triangle BCE</math> is equilateral, so <math>BC=CE=BE=7</math>. |
I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines: | I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines: |
Revision as of 14:10, 10 July 2019
Contents
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 Using Only Elementary Geometry Methods
Let be the unique point on circle
such that
is equilateral. Point
must be the point on
such that
is right. We extend line
to hit circle
again at
.
Claim 1: is the midpoint of minor arc
.
Proof: Notice that
and
are subtended by the same arc, thus they are both equal to 90 degrees.
Now it is obvious that . After this, we length chase with similar triangles and Pythag, which yields
,
,
,
, and
. We can tie this all together with Ptolemy's theorem for cyclic quadrilaterals, and we find that
. Thus,
, and the answer is
-jj_ca888
Solution 1
Use the angle bisector theorem to find ,
, and use 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 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
,
and
. Therefore,
by AA Similarity. Since side lengths in similar triangles are proportional,
However, the problem asks for , so
.
-Solution by TheBoomBox77
Solution 4
It can be verified with law of cosines that Also,
is the midpoint of major arc
so
and
Thus
is equilateral. Notice now that
But
so
bisects
Thus,
Let By law of cosines on
we find
But by ptolemy on
,
so
so
and the answer is
~abacadaea
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.