2005 AIME II Problems/Problem 12

Problem

Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$

Solutions

Solution 1 (trigonometry)

$[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^O); label("$A$",A,(-1,1));label("$B$",B,(1,1));label("$C$",C,(1,-1));label("$D$",D,(-1,-1)); label("$E$",E,(0,1));label("$F$",F,(1,1));label("$G$",G,(-1,1));label("$O$",O,(1,-1)); label("$x$",E/2+G/2,(0,1));label("$y$",G/2+F/2,(0,1)); label("$450$",(O+G)/2,(-1,1)); [/asy]$

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)

$[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("$A$",A,(-1,1));label("$B$",B,(1,1));label("$C$",C,(1,-1));label("$D$",D,(-1,-1)); label("$E$",E,(0,1));label("$F$",F,(1,1));label("$G$",G,(1,0));label("$J$",J,(1,0));label("$O$",O,(1,-1)); label("$x$",(B+F)/2,(0,1)); label("$400$",(E+F)/2,(0,1)); label("$900$",(C+D)/2,(0,-1)); [/asy]$

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,

$\begin{align*} (500-x)^2+x^2&=400^2 \\ 250000-1000x+2x^2&=16000 \\ 90000-1000x+2x^2&=0 \end{align*}$

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)

$[asy] size(3inch); pair A, B, C, D, M, O, X, Y; A = (0,900); B = (900,900); C = (900,0); D = (0,0); M = (450,900); O = (450,450); X = (250 - 50*sqrt(7),900); Y = (650 - 50*sqrt(7),900); draw(A--B--C--D--cycle); draw(X--O--Y); draw(M--O--A); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",X,N); label("$F$",Y,NNE); label("$O$",O,S); label("$M$",M,N); [/asy]$ 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 $\frac{1}{2}ab \frac{\sqrt 2}{2} = 202500$ and $a^2+b^2-ab \sqrt{2} = 160000$ Solving for $ab$ 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 $x = 250 + 50\sqrt{7}$

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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