AIME 2012 II · 第 15 题

Solution 1

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

In triangle $AEF$, let $X$ be the foot of the altitude from $A$; then $EF=EX+XF$, where we use signed lengths. Writing $EX=AE \cdot \cos \angle AEF$ and $XF=AF \cdot \cos \angle AFE$, we get

$$\begin{aligned}\tag{1} EF = AE \cdot \cos \angle AEF + AF \cdot \cos \angle AFE. \end{aligned}$$ Note $\angle AFE = \angle ACE$, and the Law of Cosines in $\triangle ACE$ gives $\cos \angle ACE = -\tfrac 17$. Also, $\angle AEF = \angle DEF$, and $\angle DFE = \tfrac{\pi}{2}$ ($DE$ is a diameter), so $\cos \angle AEF = \tfrac{EF}{DE} = \tfrac{8}{49}\cdot EF$.

Plugging in all our values into equation $(1)$, we get:

$$EF = \tfrac{64}{49} EF -\tfrac{1}{7} AF \quad \Longrightarrow \quad EF = \tfrac{7}{15} AF.$$ The Law of Cosines in $\triangle AEF$, with $EF=\tfrac 7{15}AF$ and $\cos\angle AFE = -\tfrac 17$ gives

$$8^2 = AF^2 + \tfrac{49}{225} AF^2 + \tfrac 2{15} AF^2 = \tfrac{225+49+30}{225}\cdot AF^2$$ Thus $AF^2 = \frac{900}{19}$. The answer is $\boxed{919}$. ~Shen Kislay Kai

Solution 2

Let $a = BC$, $b = CA$, $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$. We claim that $\angle MAD=\angle DAF$.

$\textit{Proof}$. Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\perp BC$. Therefore, $M\in \gamma$. Now let $X = FD\cap \omega$. Since $\angle EFX=90^\circ$, $EX$ is a diameter, so $X$ lies on the perpendicular bisector of $BC$; hence $E$, $M$, $X$ are collinear. From $\angle DAG = \angle DMX = 90^\circ$, quadrilateral $ADMX$ is cyclic. Therefore, $\angle MAD = \angle MXD$. But $\angle MXD$ and $\angle EAF$ are both subtended by arc $EF$ in $\omega$, so they are equal. Thus $\angle MAD=\angle DAF$, as claimed.

As a result, $\angle CAM = \angle FAB$. Combined with $\angle BFA=\angle MCA$, we get $\triangle ABF\sim\triangle AMC$ and therefore

$$\frac c{AM}=\frac {AF}b\qquad \Longrightarrow \qquad AF^2=\frac{b^2c^2}{AM^2} = \frac{15^2}{AM^2}$$ By Stewart's Theorem on $\triangle ABC$ (with cevian $AM$), we get

$$AM^2 = \tfrac 12 (b^2+c^2)-\tfrac 14 a^2 = \tfrac{19}{4},$$ so $AF^2 = \tfrac{900}{19}$, so the answer is $900+19=\boxed{919}$.

-Solution by thecmd999

Solution 3

Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Then use the Extended Law of Sine to find that the length of the circumradius of $\triangle ABC$ is $\tfrac{7\sqrt{3}}{3}$.

Since $DE$ is the diameter of circle $\gamma$, $\angle DFE$ is $90^\circ$. Extending $DF$ to intersect circle $\omega$ at $X$, we find that $XE$ is the diameter of $\omega$ (since $\angle DFE$ is $90^\circ$). Therefore, $XE=\tfrac{14\sqrt{3}}{3}$.

Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get

$$(u+v)^2-v^2=\frac{196}{3}-\frac{2401}{64}$$ which simplifies to

$$u^2+2uv = \frac{5341}{192}.$$ By Power of Point $D$, $uv=BD \cdot DC=735/64$. Combining with above, we get

$$XD^2=u^2=\frac{931}{192}.$$ Note that $\triangle XDE\sim \triangle ADF$ and the ratio of similarity is $\rho = AD : XD = \tfrac{15}{8}:u$. Then $AF=\rho\cdot XE = \tfrac{15}{8u}\cdot R$ and

$$AF^2 = \frac{225}{64}\cdot \frac{R^2}{u^2} = \frac{900}{19}.$$ The answer is $900+19=\boxed{919}$.

-Solution by TheBoomBox77

Solution 4

Use Law of Cosines in $\triangle ABC$ to get $\angle BAC=120^\circ$. Because $AE$ bisects $\angle A$, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\angle BEC=60^\circ.$ Thus $\triangle BEC$ is equilateral. Notice now that $\angle BFC=\angle BFE= 60^\circ.$ But $\angle DFE=90^\circ$ so $FD$ bisects $\angle BFC.$ Thus,

$$\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.$$ Let $BF=5k, CF=3k.$ Use Law of Cosines on $\triangle BFC$ to get

$$25k^2+9k^2-15k^2 = 49 \qquad \Longrightarrow\qquad k=\frac 7{\sqrt{19}}$$ Use Ptolemy's Theorem on $BFCA$, to get

$$15k+15k=7\cdot AF, \qquad \Longrightarrow\qquad AF= \frac{30}{\sqrt{19}},$$ so $AF^2=\frac{900}{19}$ and the answer is $900+19=919$

~Shen Kislay Kai

Solution 5

Denote $AB = c, BC = a, AC = b, \angle A = 2 \alpha.$ Let M be midpoint BC. Let $\theta$ be the circle centered at $A$ with radius $\sqrt{AB \cdot AC} =\sqrt{bc}.$

We calculate the length of some segments. The median $AM = \sqrt{\frac {b^2}{2} + \frac {c^2}{2} - \frac {a^2}{4}}.$ The bisector $AD = \frac {2 b c \cos \alpha}{b+c}.$ One can use Stewart's Theorem in both cases.

$AD$ is bisector of $\angle A \implies BD = \frac {a c}{b + c}, CD = \frac {a b}{b + c} \implies$

$$BD \cdot CD = \frac {a^2 bc }{(b+c)^2}.$$ We use Power of Point $D$ and get $AD \cdot DE = BD \cdot CD.$

$$AE = AD + DE = AD + \frac {BD \cdot CD}{AD},$$ $$AE =\frac {2 b c \cos \alpha}{b+c} + \frac {a^2 bc \cdot (b+c) }{(b+c)^2 \cdot 2 b c \cos \alpha} =$$ $$= \frac {b c \cos^2 \alpha + a^2}{2(b+c)\cos \alpha} =\frac {4bc \cos^2 \alpha + b^2 +c^2 -2 b c \cos 2\alpha}{2(b+c) \cos \alpha} = \frac {b+c}{2} \implies AD \cdot AE = 2 bc \cos \alpha.$$ We consider the inversion with respect $\theta.$

$B$ swap $B' \implies AB' = AC, B' \in AB \implies B'$ is symmetric to $C$ with respect to $AE.$

$C$ swap $C' \implies AC' = AB, C'$ lies on line $AC \implies C'$ is symmetric to $B$ with respect to $AE.$

$BC^2 = AB^2 + AC^2 + AB \cdot BC \implies \alpha = 60^\circ \implies AD \cdot AE = bc \implies D$ swap $E.$

Points $D$ and $E$ lies on $\Gamma \implies \Gamma$ swap $\Gamma.$

$DE$ is diameter $\Gamma, \angle DME = 90^\circ \implies M \in \Gamma.$ Therefore $M$ is crosspoint of $BC$ and $\Gamma.$

Let $\Omega$ be circumcircle $AB'C'. \Omega$ is image of line $BC.$ Point $M$ maps into $M' \implies M' = \Gamma \cap \Omega.$

Points $A, B',$ and $C'$ are symmetric to $A, C,$ and $B,$ respectively.

Point $M'$ lies on $\Gamma$ which is symmetric with respect to $AE$ and on $\Omega$ which is symmetric to $\omega$ with respect to $AE \implies$

$M'$ is symmetric $F$ with respect to $AE \implies AM' = AF.$

We use Power of Point $A$ and get

$$AF = AM' = \frac {AD \cdot AE}{AM} = \frac {4b c}{\sqrt{2 b^2 + 2 c^2 – a^2}} = \frac {4 \cdot 3 \cdot 5}{\sqrt{ 50 + 18 – 49}} = \frac {30}{\sqrt{19}} \implies \boxed{\textbf{919}}.$$ vladimir.shelomovskii@gmail.com, vvsss

Solution 6:

To do this, we first define the intersection of $EF$ and $BC$ to be $K$.

Lemma 1: $(K, C, D, B)$ are harmonic. First of all, define the midpoint of $BC$ to be $M$. Then, we have that angle FMD is $90$ degrees, and as a result, $M$ lies on this circle. By Power of a Point, $(KD)(KM) = (KE)(KF) = (KB)(KC)$. As a result, $(K, C, D, B)$ are harmonic from another famous harmonic lemma.

As a result, since $\angle EFD = 90^{\circ}$, by another Harmonic Lemma, $FD$ is the angle bisector of $BFC$. Since $\frac{BD}{CD} = \frac{5}{3}$ by angle bisector theorem, $\frac{BF}{CF} = \frac{5}{3}$. Since $\angle BAC$ is $120^{\circ}$ by Law of Cosines (LOC), we can use LOC to finish off. Call $BF = 5a$, and $CF = 3a$, $(5a)^2+(3a)^2-15a^2 = 49$, so $a = \frac{7}{\sqrt{19}}$. We do Ptolemy's Theorem on $ABFC$. Our answer is:

$$\frac{(3)(35)+(5)(21)}{7\sqrt{19}} = \frac{30}{\sqrt{19}}.$$ As a result, the final answer is $\boxed{919}$.

-sepehr2010

Minor edits ~Zhenghua

Video Solution by mop 2024

https://youtu.be/mIFUuY4ybeg

~r00tsOfUnity