Difference between revisions of "2012 AIME II Problems/Problem 15"
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Triangle <math>ABC</math> is inscribed in circle <math>\omega</math> with <math>AB=5</math>, <math>BC=7</math>, and <math>AC=3</math>. The bisector of angle <math>A</math> meets side <math>\overline{BC}</math> at <math>D</math> and circle <math>\omega</math> at a second point <math>E</math>. Let <math>\gamma</math> be the circle with diameter <math>\overline{DE}</math>. Circles <math>\omega</math> and <math>\gamma</math> meet at <math>E</math> and a second point <math>F</math>. Then <math>AF^2 = \frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Triangle <math>ABC</math> is inscribed in circle <math>\omega</math> with <math>AB=5</math>, <math>BC=7</math>, and <math>AC=3</math>. The bisector of angle <math>A</math> meets side <math>\overline{BC}</math> at <math>D</math> and circle <math>\omega</math> at a second point <math>E</math>. Let <math>\gamma</math> be the circle with diameter <math>\overline{DE}</math>. Circles <math>\omega</math> and <math>\gamma</math> meet at <math>E</math> and a second point <math>F</math>. Then <math>AF^2 = \frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | == Solution | + | ==Quick Solution using Olympiad Terms == |
− | |||
− | + | Take a force-overlaid inversion about <math>A</math> and note <math>D</math> and <math>E</math> map to each other. As <math>DE</math> was originally the diameter of <math>\gamma</math>, <math>DE</math> is still the diameter of <math>\gamma</math>. Thus <math>\gamma</math> is preserved. Note that the midpoint <math>M</math> of <math>BC</math> lies on <math>\gamma</math>, and <math>BC</math> and <math>\omega</math> are swapped. Thus points <math>F</math> and <math>M</math> map to each other, and are isogonal. It follows that <math>AF</math> is a symmedian of <math>\triangle{ABC}</math>, or that <math>ABFC</math> is harmonic. Then <math>(AB)(FC)=(BF)(CA)</math>, and thus we can let <math>BF=5x, CF=3x</math> for some <math>x</math>. By the LoC, it is easy to see <math>\angle{BAC}=120^\circ</math> so <math>(5x)^2+(3x)^2-2\cos{60^\circ}(5x)(3x)=49</math>. Solving gives <math>x^2=\frac{49}{19}</math>, from which by Ptolemy's we see <math>AF=\frac{30}{\sqrt{19}}</math>. We conclude the answer is <math>900+19=\boxed{919}</math>. | |
− | + | '''- 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. | ||
− | <math> | + | 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 <math>\sqrt{AB \cdot AC}</math> and center <math>A</math>, then reflect over the <math>A</math>-angle bisector, which fixes <math>B, C</math>). We try applying this to the problem, and it's fruitful - we end up with this solution. |
+ | -MSC | ||
− | <math> | + | == Solution 1== |
+ | 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>. 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> | ||
+ | size(150); | ||
+ | defaultpen(fontsize(9pt)); | ||
+ | picture pic; | ||
+ | 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); | ||
+ | draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic); | ||
+ | dot("$W$",circumcenter(A,B,C),dir(180)); label("$\gamma$",gamma,dir(180)); | ||
+ | </asy> | ||
+ | In triangle <math>AEF</math>, let <math>X</math> be the foot of the altitude from <math>A</math>; then <math>EF=EX+XF</math>, where we use signed lengths. Writing <math>EX=AE \cdot \cos \angle AEF</math> and <math>XF=AF \cdot \cos \angle AFE</math>, we get | ||
+ | <cmath>\begin{align}\tag{1} | ||
+ | EF = AE \cdot \cos \angle AEF + AF \cdot \cos \angle AFE. | ||
+ | \end{align}</cmath> | ||
+ | Note <math>\angle AFE = \angle ACE</math>, and the Law of Cosines in <math>\triangle ACE</math> gives <math>\cos \angle ACE = -\tfrac 17</math>. | ||
+ | Also, <math>\angle AEF = \angle DEF</math>, and <math>\angle DFE = \tfrac{\pi}{2}</math> (<math>DE</math> is a diameter), so <math>\cos \angle AEF = \tfrac{EF}{DE} = \tfrac{8}{49}\cdot EF</math>. | ||
− | + | Plugging in all our values into equation <math>(1)</math>, we get: | |
+ | <cmath>EF = \tfrac{64}{49} EF -\tfrac{1}{7} AF \quad \Longrightarrow \quad EF = \tfrac{7}{15} AF.</cmath> | ||
+ | The Law of Cosines in <math>\triangle AEF</math>, with <math>EF=\tfrac 7{15}AF</math> and <math>\cos\angle AFE = -\tfrac 17</math> gives | ||
+ | <cmath>8^2 = AF^2 + \tfrac{49}{225} AF^2 + \tfrac 2{15} AF^2 = \tfrac{225+49+30}{225}\cdot AF^2</cmath> | ||
+ | Thus <math>AF^2 = \frac{900}{19}</math>. The answer is <math>\boxed{919}</math>. | ||
− | + | == Solution 2== | |
− | <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>\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. | |
+ | <asy> | ||
+ | size(175); | ||
+ | defaultpen(fontsize(10pt)); | ||
+ | picture pic; | ||
+ | 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); pair M=MP("M",midpoint(B--C),dir(220)); | ||
− | |||
− | + | draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, gray); draw(pic, A--M--C--cycle^^A--B--F--cycle); draw(A--M, royalblue); | |
− | + | dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); | |
− | + | draw(A--B--F--cycle, black+1); | |
− | + | </asy> | |
− | + | As a result, <math>\angle CAM = \angle FAB</math>. Combined with <math>\angle BFA=\angle MCA</math>, we get <math>\triangle ABF\sim\triangle AMC</math> and therefore<cmath>\frac c{AM}=\frac {AF}b\qquad \Longrightarrow \qquad AF^2=\frac{b^2c^2}{AM^2} = \frac{15^2}{AM^2}</cmath> | |
− | + | By Stewart's Theorem on <math>\triangle ABC</math> (with cevian <math>AM</math>), we get <cmath>AM^2 = \tfrac 12 (b^2+c^2)-\tfrac 14 a^2 = \tfrac{19}{4},</cmath> so <math>AF^2 = \tfrac{900}{19}</math>, so the answer is <math>900+19=\boxed{919}</math>. | |
'''-Solution by thecmd999''' | '''-Solution by thecmd999''' | ||
==Solution 3== | ==Solution 3== | ||
+ | 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 Extended Law of Sine to find that the length of the circumradius of <math>\triangle ABC</math> is <math>\tfrac{7\sqrt{3}}{3}</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>. | ||
− | < | + | 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>XD^2=u^2=\frac{931}{192}.</cmath> | ||
+ | Note that <math>\triangle XDE\sim \triangle ADF</math> and the ratio of similarity is <math>\rho = AD : XD = \tfrac{15}{8}:u</math>. Then <math>AF=\rho\cdot XE = \tfrac{15}{8u}\cdot R</math> and <cmath>AF^2 = \frac{225}{64}\cdot \frac{R^2}{u^2} = \frac{900}{19}.</cmath> | ||
+ | The answer is <math>900+19=\boxed{919}</math>. | ||
− | + | '''-Solution by TheBoomBox77''' | |
− | <cmath> | + | ==Solution 4== |
+ | Use Law of Cosines in <math>\triangle ABC</math> to get <math>\angle BAC=120^\circ</math>. Because <math>AE</math> bisects <math>\angle A</math>, <math>E</math> is the midpoint of major arc <math>BC</math> so <math>BE=CE,</math> and <math>\angle BEC=60^\circ.</math> Thus <math>\triangle BEC</math> is equilateral. Notice now that <math>\angle BFC=\angle BFE= 60^\circ.</math> But <math>\angle DFE=90^\circ</math> so <math>FD</math> bisects <math>\angle BFC.</math> Thus, <cmath>\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.</cmath> | ||
+ | Let <math>BF=5k, CF=3k.</math> Use Law of Cosines on <math>\triangle BFC</math> to get<cmath>25k^2+9k^2-15k^2 = 49 \qquad \Longrightarrow\qquad k=\frac 7{\sqrt{19}}</cmath> Use Ptolemy's Theorem on <math>BFCA</math>, to get <cmath>15k+15k=7\cdot AF, \qquad \Longrightarrow\qquad AF= \frac{30}{\sqrt{19}},</cmath> so <math>AF^2=\frac{900}{19}</math> and the answer is <math>900+19=919</math> | ||
− | + | ~abacadaea | |
− | < | + | ==Solution 5== |
+ | [[File:2012 AIME II 15a.png|500px|right]] | ||
+ | Denote <math>AB = c, BC = a, AC = b, \angle A = 2 \alpha.</math> | ||
+ | Let M be midpoint BC. Let <math>\theta</math> be the circle centered at <math>A</math> with radius <math>\sqrt{AB \cdot AC} =\sqrt{bc}.</math> | ||
− | <cmath>a^2+ | + | We calculate the length of some segments. |
+ | The median <math>AM = \sqrt{\frac {b^2}{2} + \frac {c^2}{2} - \frac {a^2}{4}}.</math> | ||
+ | The bisector <math>AD = \frac {2 b c \cos \alpha}{b+c}.</math> One can use Stewart's Theorem in both cases. | ||
+ | |||
+ | <math>AD</math> is bisector of <math>\angle A \implies BD = \frac {a c}{b + c}, CD = \frac {a b}{b + c} \implies</math> | ||
+ | <cmath>BD \cdot CD = \frac {a^2 bc }{(b+c)^2}.</cmath> | ||
+ | We use Power of Point <math>D</math> and get <math>AD \cdot DE = BD \cdot CD. </math> | ||
+ | <cmath>AE = AD + DE = AD + \frac {BD \cdot CD}{AD},</cmath> | ||
+ | <cmath>AE =\frac {2 b c \cos \alpha}{b+c} + \frac {a^2 bc \cdot (b+c) }{(b+c)^2 \cdot 2 b c \cos \alpha} =</cmath> | ||
+ | <cmath>= \frac {b c \cos^2 \alpha + a^2}{2(b+c)\cos \alpha} =\frac {4bc \cos^2 \alpha + b^2 +c^2 -2 b c \cos 2\alpha}{2(b+c) \cos \alpha} = \frac {b+c}{2} \implies AD \cdot AE = 2 bc \cos \alpha.</cmath> | ||
+ | We consider the inversion with respect <math>\theta.</math> | ||
− | + | <math>B</math> swap <math>B' \implies AB' = AC, B' \in AB \implies B'</math> is symmetric to <math>C</math> with respect to <math>AE.</math> | |
− | < | + | <math>C</math> swap <math>C' \implies AC' = AB, C'</math> lies on line <math>AC \implies C'</math> is symmetric to <math>B</math> with respect to <math>AE.</math> |
− | < | + | <math>BC^2 = AB^2 + AC^2 + AB \cdot BC \implies \alpha = 60^\circ \implies AD \cdot AE = bc \implies D</math> swap <math>E.</math> |
− | + | Points <math>D</math> and <math>E</math> lies on <math>\Gamma \implies \Gamma </math> swap <math>\Gamma.</math> | |
− | + | <math>DE</math> is diameter <math>\Gamma, \angle DME = 90^\circ \implies M \in \Gamma.</math> Therefore <math>M</math> is crosspoint of <math>BC</math> and <math>\Gamma.</math> | |
− | < | + | Let <math>\Omega</math> be circumcircle <math>AB'C'. \Omega</math> is image of line <math>BC.</math> |
+ | Point <math>M</math> maps into <math>M' \implies M' = \Gamma \cap \Omega.</math> | ||
− | < | + | Points <math>A, B',</math> and <math>C'</math> are symmetric to <math>A, C,</math> and <math>B,</math> respectively. |
− | < | + | Point <math>M'</math> lies on <math>\Gamma</math> which is symmetric with respect to <math>AE</math> and on <math>\Omega</math> which is symmetric to <math>\omega</math> with respect to <math>AE \implies</math> |
− | < | + | <i><b><math>M'</math> is symmetric <math>F</math> with respect to <math>AE \implies AM' = AF.</math></b></i> |
− | + | We use Power of Point <math>A</math> and get | |
+ | <cmath>AF = AM' = \frac {AD \cdot AE}{AM} = \frac {4b c}{\sqrt{2 b^2 + 2 c^2 – a^2}} = \frac {4 \cdot 3 \cdot 5}{\sqrt{ 50 + 18 – 49}} | ||
+ | = \frac {30}{\sqrt{19}} \implies \boxed{\textbf{919}}.</cmath> | ||
− | ''' | + | '''vladimir.shelomovskii@gmail.com, vvsss''' |
− | ==Solution | + | ==Video Solution by mop 2024== |
− | + | https://youtu.be/mIFUuY4ybeg | |
− | + | ~r00tsOfUnity | |
− | |||
− | ~ | ||
== 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}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:56, 10 September 2023
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 Extended Law of Sine 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
(since
is
). Therefore,
.
Let ,
, and
. Then
, so we get
which simplifies to
By Power of Point
,
. Combining with above, we get
Note that
and the ratio of similarity is
. Then
and
The answer is
.
-Solution by TheBoomBox77
Solution 4
Use Law of Cosines in to get
. Because
bisects
,
is the midpoint of major arc
so
and
Thus
is equilateral. Notice now that
But
so
bisects
Thus,
Let
Use Law of Cosines on
to get
Use Ptolemy's Theorem on
, to get
so
and the answer is
~abacadaea
Solution 5
Denote
Let M be midpoint BC. Let
be the circle centered at
with radius
We calculate the length of some segments.
The median
The bisector
One can use Stewart's Theorem in both cases.
is bisector of
We use Power of Point
and get
We consider the inversion with respect
swap
is symmetric to
with respect to
swap
lies on line
is symmetric to
with respect to
swap
Points and
lies on
swap
is diameter
Therefore
is crosspoint of
and
Let be circumcircle
is image of line
Point
maps into
Points and
are symmetric to
and
respectively.
Point lies on
which is symmetric with respect to
and on
which is symmetric to
with respect to
is symmetric
with respect to
We use Power of Point and get
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by mop 2024
~r00tsOfUnity
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.