Difference between revisions of "2012 AIME II Problems/Problem 15"
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+ | Use the angle bisector theorem to find <math>CD=\tfrac{21}{8}</math>, <math>BD=\tfrac{35}{8}</math>, and use Stewart's Theorem to find <math>AD=\tfrac{15}{8}</math>. Use Power of Point <math>D</math> to find <math>DE=\tfrac{49}{8}</math>, and so <math>AE=8</math>. Then use the <math>\textit{extended law of sines}</math> to find that the length of the circumradius of <math>\triangle ABC</math> is <math>\tfrac{7\sqrt{3}}{3}</math>. Use law of cosines to find <math>\angle CAD = \tfrac{\pi} {3}</math>, hence <math>\angle BAD = \tfrac{\pi}{3}</math> as well, and <math>\triangle BCE</math> is equilateral, so <math>BC=CE=BE=7</math>. | ||
<asy> | <asy> | ||
− | size( | + | size(175); |
− | pair | + | defaultpen(fontsize(9pt)); |
− | + | pair A,B,C,D,E,F,W; | |
− | + | B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); | |
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− | + | draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); | |
− | + | dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); label("$u$",X--D,dir(60)); label("$v$",D--F,dir(70)); | |
− | + | </asy> | |
− | + | Since <math>DE</math> is the diameter of circle <math>\gamma</math>, <math>\angle DFE</math> is <math>90^\circ</math>. Extending <math>DF</math> to intersect circle <math>\omega</math> at <math>X</math>, we find that <math>XE</math> is the diameter of <math>\omega</math> (since <math>\angle DFE</math> is <math>90^\circ</math>). Therefore, <math>XE=\tfrac{14\sqrt{3}}{3}</math>. | |
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− | + | Let <math>EF=x</math>, <math>XD=u</math>, and <math>DF=v</math>. Then <math>XE^2-XF^2=EF^2=DE^2-DF^2</math>, so we get | |
+ | <cmath>(u+v)^2-v^2=\frac{196}{3}-\frac{2401}{64}</cmath> | ||
+ | which simplifies to | ||
+ | <cmath>u^2+2uv = \frac{5341}{192}.</cmath> | ||
+ | By Power of Point <math>D</math>, <math>uv=BD \cdot DC=735/64</math>. Combining with above, we get | ||
+ | <cmath>u^2=\frac{931}{192}\qquad \Longrightarrow\qquad XD=u=\frac{7\sqrt{19}}{8\sqrt{3}}.</cmath> | ||
+ | Because <math>\angle EXF</math> and <math>\angle EAF</math> intercept the same arc in circle <math>\omega</math>, <math>\angle EXF = \angle EAF</math>; similarly, <math>\angle XFA =\angle XEA</math>. Therefore, <math>\triangle XDE\sim\triangle ADF</math>. We get <math>AF:XE = AD:XD</math>, i.e., <math>AF\cdot u = AD\cdot XE</math>. Thus | ||
+ | <cmath>AF \cdot \frac{7\sqrt{19}}{8\sqrt{3}} = \frac{15}{8} \cdot \frac{14\sqrt{3}}{3} \qquad \Longrightarrow\qquad AF = \frac{30}{\sqrt{19}}.</cmath> | ||
+ | Thus <math>AF^2 = \tfrac{900}{19}</math>, so the answer is <math>900+19=\boxed{919}</math>. | ||
'''-Solution by TheBoomBox77''' | '''-Solution by TheBoomBox77''' |
Revision as of 02:55, 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
Use the angle bisector theorem to find ,
, and use Stewart's Theorem to find
. Use Power of Point
to find
, and so
. Then use the
to find that the length of the circumradius of
is
. Use law of cosines to find
, hence
as well, and
is equilateral, so
.
Since
is the diameter of circle
,
is
. Extending
to intersect circle
at
, we find that
is the diameter of
(since
is
). Therefore,
.
Let ,
, and
. Then
, so we get
which simplifies to
By Power of Point
,
. Combining with above, we get
Because
and
intercept the same arc in circle
,
; similarly,
. Therefore,
. We get
, i.e.,
. Thus
Thus
, so the answer is
.
-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.