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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>.
  
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== Solution ==
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== Solution 1==
 
Use the angle bisector theorem to find <math>CD=21/8</math>, <math>BD=35/8</math>, and use the Stewart's Theorem to find <math>AD=15/8</math>.  Use Power of the Point to find <math>DE=49/8</math>, and so <math>AE=8</math>.  Use law of cosines to find <math>\angle CAD = \pi /3</math>, hence <math>\angle BAD = \pi /3</math> as well, and <math>\triangle BCE</math> is equilateral, so <math>BC=CE=BE=7</math>.
 
Use the angle bisector theorem to find <math>CD=21/8</math>, <math>BD=35/8</math>, and use the Stewart's Theorem to find <math>AD=15/8</math>.  Use Power of the Point to find <math>DE=49/8</math>, and so <math>AE=8</math>.  Use law of cosines to find <math>\angle CAD = \pi /3</math>, hence <math>\angle BAD = \pi /3</math> as well, and <math>\triangle BCE</math> is equilateral, so <math>BC=CE=BE=7</math>.
  
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<math>64 = \frac{AF^2}{225} \cdot (225+49+30),</math> or <math>AF^2 = \frac{900}{19}.</math>  The answer is <math>\boxed{919}</math>.
 
<math>64 = \frac{AF^2}{225} \cdot (225+49+30),</math> or <math>AF^2 = \frac{900}{19}.</math>  The answer is <math>\boxed{919}</math>.
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== Solution 2==
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Let <math>a = BC</math>, <math>b = CA</math>, <math>c = AB</math> for convenience. We claim that <math>AF</math> is a symmedian. Indeed, let <math>M</math> be the midpoint of segment <math>BC</math>. Since <math>\angle EAB=\angle EAC</math>, it follows that <math>EB = EC</math> and consequently <math>EM\perp BC</math>. Therefore, <math>M\in \gamma</math>. Now let <math>G = FD\cap \omega</math>. Since <math>EG</math> is a diameter, <math>G</math> lies on the perpendicular bisector of <math>BC</math>; hence <math>E</math>, <math>M</math>, <math>G</math> are collinear. From <math>\angle DAG = \angle DMG = 90</math>, it immediately follows that quadrilateral <math>ADMG</math> is cyclic. Therefore, <math>\angle MAD = \angle MGD=\angle EAF</math>, implying that <math>AF</math> is a symmedian, as claimed.
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The rest is standard; here's a quick way to finish. From above, quadrilateral <math>ABFC</math> is harmonic, so <math>\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}</math>. In conjunction with <math>\triangle ABF\sim\triangle AMC</math>, it follows that <math>AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}</math>. (Notice that this holds for all triangles <math>ABC</math>.) To finish, substitute <math>a = 7</math>, <math>b=3</math>, <math>c=5</math> to obtain <math>AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}</math> as before.
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'''-Solution by thecmd999'''
  
 
== 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 00:21, 12 March 2014

Problem 15

Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

Use the angle bisector theorem to find $CD=21/8$, $BD=35/8$, and use the Stewart's Theorem to find $AD=15/8$. Use Power of the Point to find $DE=49/8$, and so $AE=8$. Use law of cosines to find $\angle CAD = \pi /3$, hence $\angle BAD = \pi /3$ as well, and $\triangle BCE$ is equilateral, so $BC=CE=BE=7$.

I'm sure there is a more elegant solution from here, but instead we do'll some hairy law of cosines:

$AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot cos \angle AFE.$ (1)

$AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot cos \angle AEF.$ Adding these two and simplifying we get:

$EF = AF \cdot cos \angle AFE + AE \cdot cos \angle AEF$ (2). Ah, but $\angle AFE = \angle ACE$ (since $F$ lies on $\omega$), and we can find $cos \angle ACE$ using the law of cosines:

$AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot cos \angle ACE$, and plugging in $AE = 8, AC = 3, BE = BC = 7,$ we get $cos \angle ACE = -1/7 = cos \angle AFE$.

Also, $\angle AEF = \angle DEF$, and $\angle DFE = \pi/2$ (since $F$ is on the circle $\gamma$ with diameter $DE$), so $cos \angle AEF = EF/DE = 8 \cdot EF/49$.

Plugging in all our values into equation (2), we get:

$EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}$, or $EF = \frac{7}{15} \cdot AF$.

Finally, we plug this into equation (1), yielding:

$8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}$. Thus,

$64 = \frac{AF^2}{225} \cdot (225+49+30),$ or $AF^2 = \frac{900}{19}.$ The answer is $\boxed{919}$.

Solution 2

Let $a = BC$, $b = CA$, $c = AB$ for convenience. We claim that $AF$ is a symmedian. Indeed, let $M$ be the midpoint of segment $BC$. Since $\angle EAB=\angle EAC$, it follows that $EB = EC$ and consequently $EM\perp BC$. Therefore, $M\in \gamma$. Now let $G = FD\cap \omega$. Since $EG$ is a diameter, $G$ lies on the perpendicular bisector of $BC$; hence $E$, $M$, $G$ are collinear. From $\angle DAG = \angle DMG = 90$, it immediately follows that quadrilateral $ADMG$ is cyclic. Therefore, $\angle MAD = \angle MGD=\angle EAF$, implying that $AF$ is a symmedian, as claimed.

The rest is standard; here's a quick way to finish. From above, quadrilateral $ABFC$ is harmonic, so $\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}$. In conjunction with $\triangle ABF\sim\triangle AMC$, it follows that $AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}$. (Notice that this holds for all triangles $ABC$.) To finish, substitute $a = 7$, $b=3$, $c=5$ to obtain $AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}$ as before.

-Solution by thecmd999

See Also

2012 AIME II (Problems • Answer Key • Resources)
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