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2005 AIME II Problems/Problem 12

Revision as of 00:50, 19 June 2020 by Jerry guo (talk | contribs) (→‎Solution 7 (Using a Circle))

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

[asy] size(250); pair A,B,C,D,O,E,F,G,H,K; A = (0,0); B = (900,0); C = (900,900); D = (0,900); O = (450,450); E = (600,0); F = (150,0); G = (-600,0); H = (450,0); K = (0,270); draw(A--B--C--D--cycle); draw(O--E); draw(O--F); draw(O--G); draw(A--G); draw(O--H); label("O",O,N); label("A",A,S); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("E",E,SE); label("F",F,S); label("H",H,SW); label("G",G,SW); label("x",H--E,S); label("K",K,NW); [/asy] 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)

[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=(4.5-0.5*sqrt(7),7), O=(4.5,4.5); draw(A--B--C--D--A); draw(E--O--F); draw(E--G--F); draw(circle(G,2*sqrt(2))); markscalefactor=0.1; draw(rightanglemark(E,G,F)); 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,(0,-1)); label("\(O\)",O,(1,-1)); [/asy]

See also

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
All AIME Problems and Solutions

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