AIME 2005 II · 第 12 题

Solutions

Solution 1 (trigonometry)

Let $G$ be the foot of the perpendicular from $O$ to $AB$. Denote $x = EG$ and $y = FG$, and $x > y$ (since $AE < BF$ and $AG = BG$). Then $\tan \angle EOG = \frac{x}{450}$, and $\tan \angle FOG = \frac{y}{450}$.

By the tangent addition rule $\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)$, we see that

$$\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.$$ Since $\tan 45 = 1$, this simplifies to $1 - \frac{xy}{450^2} = \frac{x + y}{450}$. We know that $x + y = 400$, so we can substitute this to find that $1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2$.

Substituting $x = 400 - y$ again, we know have $xy = (400 - y)y = 150^2$. This is a quadratic with roots $200 \pm 50\sqrt{7}$. Since $y < x$, use the smaller root, $200 - 50\sqrt{7}$.

Now, $BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}$. The answer is $250 + 50 + 7 = \boxed{307}$.

Solution 2 (synthetic)

Label $BF=x$, so $EA =$ $500 - x$. Rotate $\triangle{OEF}$ about $O$ until $EF$ lies on $BC$. Now we know that $\angle{EOF}=45^\circ$ therefore $\angle BOF+\angle AOE=45^\circ$ also since $O$ is the center of the square. Label the new triangle that we created $\triangle OGJ$. Now we know that rotation preserves angles and side lengths, so $BG=500-x$ and $JC=x$. Draw $GF$ and $OB$. Notice that $\angle BOG =\angle OAE$ since rotations preserve the same angles so $\angle{FOG}=45^\circ$ too. By SAS we know that $\triangle FOE\cong \triangle FOG,$ so $FG=400$. Now we have a right $\triangle BFG$ with legs $x$ and $500-x$ and hypotenuse $400$. By the Pythagorean Theorem,

$$\begin{aligned} (500-x)^2+x^2&=400^2 \\ 250000-1000x+2x^2&=16000 \\ 90000-1000x+2x^2&=0 \end{aligned}$$ and applying the quadratic formula we get that $x=250\pm 50\sqrt{7}$. Since $BF > AE,$ we take the positive root, and our answer is $p+q+r = 250 + 50 + 7 = 307$.

Solution 3 (similar triangles)

Let the midpoint of $\overline{AB}$ be $M$ and let $FB = x$, so then $MF = 450 - x$ and $AF = 900 - x$. Drawing $\overline{AO}$, we have $\triangle OEF\sim\triangle AOF$, so

$$\frac{OF}{EF} = \frac{AF}{OF}\Rightarrow (OF)^2 = 400(900 - x).$$ By the Pythagorean Theorem on $\triangle OMF$,

$$(OF)^2 = 450^2 + (450 - x)^2.$$ Setting these two expressions for $(OF)^2$ equal and solving for $x$ (it is helpful to scale the problem down by a factor of 50 first), we get $x = 250\pm 50\sqrt{7}$. Since $BF > AE$, we want the value $x = 250 + 50\sqrt{7}$, and the answer is $250 + 50 + 7 = \boxed{307}$.

Solution 4 (Abusing Stewart)

Let $x = BF$, so $AE = 500-x$. Let $a = OE$, $b = OF$. Applying Stewart's Theorem on triangles $AOB$ twice, first using $E$ as the base point and then $F$, we arrive at the equations

$$(450 \sqrt{2})^2 (900) = 900(500-x)(400+x) + a^2 (900)$$ and

$$(450 \sqrt{2})^2 (900) = 900x(900-x) + b^2 (900)$$ Now applying law of sines and law of cosines on $\triangle EOF$ yields

$$\frac{1}{2} ab \sin 45^{\circ} = \frac{4}{9} \times \frac{1}{4} \times 900^2 = 202500$$ and

$$a^2+b^2- 2 ab \cos 45^{\circ} = 160000$$ Solving for $ab$ from the sines equation and plugging into the law of cosines equation yields $a^2+b^2 = 290000$. We now finish by adding the two original stewart equations and obtaining:

$$2(450\sqrt{2})^2 = (500-x)(400+x)+x(900-x)+520000$$ This is a quadratic which only takes some patience to solve for $x = 250 + 50\sqrt{7}$

Solution 5 (Complex Numbers)

Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with $o = 0, a = -450 + 450i, b = 450 + 450i$, and $f = x + 450i$. Since $EF$ = 400, $e = (x-400) + 450i$. From $\angle{EOF} = 45^{\circ}$, we can deduce that the rotation of point $F$ 45 degrees counterclockwise, $E$, and the origin are collinear. In other words,

$$\dfrac{e^{i \frac{\pi}{4}} \cdot (x + 450i)}{(x - 400) + 450i}$$ is a real number. Simplyfying using the fact that $e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}$, clearing the denominator, and setting the imaginary part equal to $0$, we eventually get the quadratic

$$x^2 - 400x + 22500 = 0$$ which has solutions $x = 200 \pm 50\sqrt{7}$. It is given that $AE < BF$, so $x = 200 - 50\sqrt{7}$ and

$$BF = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7} \Rightarrow \boxed{307}.$$ -MP8148

Solution 6

Let G be a point such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB. Since $\triangle GOE \sim \triangle OHE$, $\frac{GO}{OE} = \frac{450}{x}$, and by Angle Bisector Theorem, $\frac{GF}{FE} = \frac{450}{x}$. Thus, $GF = \frac{450 \cdot 400}{x}$. $AF = AH-FH = 50+x$, and $KA = EB$ (90 degree rotation), and now we can bash on 2 similar triangles $\triangle GAK \sim \triangle GHO$.

$$\frac{GA}{AK} = \frac{GH}{OH}$$ $$\frac{\frac{450 \cdot 400}{x}-50-x}{450-x} = \frac{\frac{450 \cdot 400}{x}+400-x}{450}$$ I hope you like expanding

$$x^2 - 850x + \frac{81000000}{x} = -450x - 22500 + \frac{81000000}{x}$$ $$x^2 - 400x + 22500 = 0$$ Quadratic formula gives us

$$x = 200 \pm 50 \sqrt{7}$$ Since AE < BF

$$x = 200 - 50 \sqrt{7}$$ Thus,

$$BF = 250 + 50 \sqrt{7}$$ So, our answer is $\boxed{307}$.

-AlexLikeMath

Solution 7 (Using a Circle)

We know that G is on the perpendicular bisector of $EF$, which means that $EJ=JF=200$, $EG=GF=200\sqrt{2}$ and $GH=250$. Now, let $HO$ be equal to $x$. We can set up an equation with the Pythagorean Theorem:

$$\begin{aligned} \sqrt{x^2+250^2}&=(200\sqrt{2})^2 \\ x^2+62500&=80000 \\ x^2&=17500 \\ x&=50\sqrt{7} \end{aligned}$$ Now, since $IO=450$,

$$\begin{aligned} HI&=450-x \\ &=450-50\sqrt{7} \\ \end{aligned} \\$$ Since $HI=AJ$, we now have:

$$\begin{aligned} BF&=AB-AJ-JF \\ &=900-(450-50\sqrt{7})-200 \\ &=250+50\sqrt{7} \\ \end{aligned}$$ This means that our answer would be $250+50+7=\boxed{307}$

~Jerry_Guo

Solution 8 (More Similar Triangles)

Construct $BO, AO.$ Let $\angle{FOB} = \alpha.$ Also let $FB = x$ then $AE = 500-x.$ We then have from simple angle-chasing:

$$\begin{aligned} \angle{BFO} = 135 - \alpha \\ \angle{OFE} = 45 + \alpha \\ \angle{EOA} = 45 - \alpha \\ \angle{AEO} = 90 + \alpha \\ \angle{OEF} = 90 - \alpha. \end{aligned}$$ From AA similarity we have

$$\triangle{EOB} \sim \triangle{EFO}.$$ This gives the ratios,

$$\dfrac{400 + x}{EO} = \dfrac{450\sqrt{2}}{FO}.$$ Similarly from AA similarity

$$\triangle{FOA} \sim \triangle{FEO}.$$ So we get the ratios

$$\dfrac{EO}{450\sqrt{2}} = \dfrac{FO}{900-x}.$$ We can multiply to get

$$\dfrac{400 + x}{450\sqrt{2}} = \dfrac{450\sqrt{2}}{900 - x}.$$ Cross-multiplying reveals

$$360000 + 500x - x^2 = 405000.$$ Bringing everything to one side we have

$$x^2 - 500x + 45000 = 0.$$ By the quadratic formula we get

$$x = \dfrac{500 + \sqrt{500^2 - 4\cdot45000}}{2} = \dfrac{500 + \sqrt{70000}}{2} = \dfrac{500 + 100\sqrt{7}}{2} = 250 + 50\sqrt{7}.$$ Therefore

$$p+q+r = 250 + 50 + 7 = \boxed{307}.$$ ~aa1024

Solution 9

We use ratio lemma and Stewart's theorem: Connect $OA, OE, OF, OB$ and let $AE = x$ and $BF = 500 - x.$ Let angle $AOE = y,$ hence $BOF = 45 - y.$ Now, we apply Stewart's theorem in triangles $AOF$ and $BOE$ to get $OE$ and $OF$ in terms of $x$ finally, calculate $x/400$ and $500-x/400$ using ratio lemma to find $x$ and $y$

Solution 10(Similar Triangles)

Draw AO, OB, and extend OB to D. Let $\angle{FOB} = \alpha.$ Then, after angle chasing, we find that

$$\angle{AEB} = 90 + \alpha$$ . Using this, we draw a line perpendicular to $AB$ at $E$ to meet $BD$ at $M$. Since $\angle{MEO} = \alpha$ and $\angle{EMO} = 45$, we have that

$$\triangle{EMO} \sim \triangle{OBF}$$ Let $FB = x$. Then $EM = 400+x$. Since $FB/BO = \frac{x}{450\sqrt{2}}$, and $MO/EM = FB/OB$, we have

$$MO = \frac{(400+x)x}{450\sqrt{2}}$$ Since $\triangle{EBM}$ is a $45-45-90$ triangle,

$$(400+x)\sqrt{2} = 450 \sqrt{2} + \frac{(400+x)x}{450\sqrt{2}}$$ Solving for $x$, we get that $x=250 +- 50s\sqrt{7}$, but since $FB>AE$, $FB = 250+50\sqrt{7}$, thus

$$p+q+r=\boxed{307}$$ -dchang0524