2005 AIME II Problems/Problem 12

Revision as of 22:06, 19 November 2016 by Ramanan369 (talk | contribs) (Solution 2 (synthetic))

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] 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

[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 $250 + 50 + 7 = \boxed{307}$.

See also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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