Difference between revisions of "2012 AIME II Problems/Problem 15"
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Let <math>a = BC</math>, <math>b = CA</math>, <math>c = AB</math> for convenience. Let <math>M</math> be the midpoint of segment <math>BC</math>. We claim that <math>\angle MAD=\angle DAF</math>. | Let <math>a = BC</math>, <math>b = CA</math>, <math>c = AB</math> for convenience. Let <math>M</math> be the midpoint of segment <math>BC</math>. We claim that <math>\angle MAD=\angle DAF</math>. | ||
− | <math>\textit{Proof}</math>. Since <math>AE</math> is the angle bisector, it follows that <math>EB = EC</math> and consequently <math>EM\perp BC</math>. Therefore, <math>M\in \gamma</math>. Now let <math>X = FD\cap \omega</math>. Since <math>EX</math> is a diameter, <math>X</math> lies on the perpendicular bisector of <math>BC</math>; hence <math>E</math>, <math>M</math>, <math>X</math> are collinear. From <math>\angle DAG = \angle DMX = 90^\circ</math>, quadrilateral <math>ADMX</math> is cyclic. Therefore, <math>\angle MAD = \angle MXD</math>. But <math>\angle MXD</math> and <math>\angle EAF</math> are both subtended by arc <math>EF</math> in <math>\omega</math>, so they are equal. Thus <math>\angle MAD=\angle DAF</math>, as claimed. | + | <math>\textit{Proof}</math>. Since <math>AE</math> is the angle bisector, it follows that <math>EB = EC</math> and consequently <math>EM\perp BC</math>. Therefore, <math>M\in \gamma</math>. Now let <math>X = FD\cap \omega</math>. Since <math>\angle EFX=90^\circ</math>, <math>EX</math> is a diameter, so <math>X</math> lies on the perpendicular bisector of <math>BC</math>; hence <math>E</math>, <math>M</math>, <math>X</math> are collinear. From <math>\angle DAG = \angle DMX = 90^\circ</math>, quadrilateral <math>ADMX</math> is cyclic. Therefore, <math>\angle MAD = \angle MXD</math>. But <math>\angle MXD</math> and <math>\angle EAF</math> are both subtended by arc <math>EF</math> in <math>\omega</math>, so they are equal. Thus <math>\angle MAD=\angle DAF</math>, as claimed. |
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Revision as of 00:37, 15 January 2022
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
.
Quick Solution using Olympiad Terms
Take a force-overlaid inversion about and note
and
map to each other. As
was originally the diameter of
,
is still the diameter of
. Thus
is preserved. Note that the midpoint
of
lies on
, and
and
are swapped. Thus points
and
map to each other, and are isogonal. It follows that
is a symmedian of
, or that
is harmonic. Then
, and thus we can let
for some
. By the LoC, it is easy to see
so
. Solving gives
, from which by Ptolemy's we see
. We conclude the answer is
.
- Emathmaster
Side Note: You might be wondering what the motivation for this solution is. Most of the people who've done EGMO Chapter 8 should recognize this as problem 8.32 (2009 Russian Olympiad) with the computational finish afterwards. Now if you haven't done this, but still know what inversion is, here's the motivation. We'd see that it's kinda hard to angle chase, and if we could, it would still be a bit hard to apply (you could use trig, but it won't be so clean most likely). If you give up after realizing that angle chasing won't work, you'd likely go in a similar approach to Solution 1 (below) or maybe be a bit more insightful and go with the elementary solution above.
Finally, we notice there's circles! Classic setup for inversion! Since we're involving an angle-bisector, the first thing that comes to mind is a force overlaid inversion described in Lemma 8.16 of EGMO (where we invert with radius and center
, then reflect over the
-angle bisector, which fixes
). We try applying this to the problem, and it's fruitful - we end up with this solution.
-MSC
Solution 1
Use the angle bisector theorem to find ,
, and use Stewart's Theorem to find
. Use Power of Point
to find
, and so
. Use law of cosines to find
, hence
as well, and
is equilateral, so
.
In triangle
, let
be the foot of the altitude from
; then
, where we use signed lengths. Writing
and
, we get
Note
, and the Law of Cosines in
gives
.
Also,
, and
(
is a diameter), so
.
Plugging in all our values into equation , we get:
The Law of Cosines in
, with
and
gives
Thus
. The answer is
.
Solution 2
Let ,
,
for convenience. Let
be the midpoint of segment
. We claim that
.
. Since
is the angle bisector, it follows that
and consequently
. Therefore,
. Now let
. Since
,
is a diameter, so
lies on the perpendicular bisector of
; hence
,
,
are collinear. From
, quadrilateral
is cyclic. Therefore,
. But
and
are both subtended by arc
in
, so they are equal. Thus
, as claimed.
As a result,
. Combined with
, we get
and therefore
By Stewart's Theorem on
(with cevian
), we get
so
, so the answer is
.
-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.