Difference between revisions of "2005 AIME II Problems/Problem 12"

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[[Square]] <math>ABCD </math> has [[center]] <math> O,\ AB=900,\ E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF </math> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r}, </math> where <math> p,q, </math> and <math> r </math> are [[positive]] [[integer]]s and <math> r </math> is not divisible by the [[square]] of any [[prime]], find <math> p+q+r. </math>
 
[[Square]] <math>ABCD </math> has [[center]] <math> O,\ AB=900,\ E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF </math> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r}, </math> where <math> p,q, </math> and <math> r </math> are [[positive]] [[integer]]s and <math> r </math> is not divisible by the [[square]] of any [[prime]], find <math> p+q+r. </math>
 
__TOC__
 
__TOC__
== Solution ==
+
 
=== Solution 1 ===
+
== Solutions ==
 +
=== Solution 1 (trigonometry) ===
 
<center><asy>
 
<center><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));  
 
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></center> <!-- Asymptote replacement for Image:AIME_2005II_Solution_12_1.png by Minsoens -->
 
</asy></center> <!-- Asymptote replacement for Image:AIME_2005II_Solution_12_1.png by Minsoens -->
Line 16: Line 18:
 
Now, <math>BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}</math>. The answer is <math>250 + 50 + 7 = \boxed{307}</math>.
 
Now, <math>BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}</math>. The answer is <math>250 + 50 + 7 = \boxed{307}</math>.
  
=== Solution 2 ===
+
=== Solution 2 (synthetic) ===
 
<center><asy>
 
<center><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));
 
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></center>
 
</asy></center>
  
 
Label <math>BF=x</math>, so <math>EA =</math> <math>500 - x</math>. Rotate <math>\triangle{OEF}</math> about <math>O</math> until <math>EF</math> lies on <math>BC</math>. Now we know that <math>\angle{EOF}=45^\circ</math> therefore <math>\angle BOF+\angle AOE=45^\circ</math> also since <math>O</math> is the center of the square.  Label the new triangle that we created <math>\triangle OGJ</math>.  Now we know that rotation preserves angles and side lengths, so <math>BG=500-x</math> and <math>JC=x</math>.  Draw <math>GF</math> and <math>OB</math>. Notice that <math>\angle BOG =\angle OAE</math> since rotations preserve the same angles so  
 
Label <math>BF=x</math>, so <math>EA =</math> <math>500 - x</math>. Rotate <math>\triangle{OEF}</math> about <math>O</math> until <math>EF</math> lies on <math>BC</math>. Now we know that <math>\angle{EOF}=45^\circ</math> therefore <math>\angle BOF+\angle AOE=45^\circ</math> also since <math>O</math> is the center of the square.  Label the new triangle that we created <math>\triangle OGJ</math>.  Now we know that rotation preserves angles and side lengths, so <math>BG=500-x</math> and <math>JC=x</math>.  Draw <math>GF</math> and <math>OB</math>. Notice that <math>\angle BOG =\angle OAE</math> since rotations preserve the same angles so  
<math>\angle{FOG}=45^\circ</math> too and by SAS we know that <math>\triangle FOE\cong \triangle FOG</math> so <math>FG=400</math>. Now we have a right <math>\triangle BFG</math> with legs <math>x</math> and <math>500-x</math> and hypotenuse <math>400</math>. Then by the [[Pythagorean Theorem]],  
+
<math>\angle{FOG}=45^\circ</math> too. By SAS we know that <math>\triangle FOE\cong \triangle FOG,</math> so <math>FG=400</math>. Now we have a right <math>\triangle BFG</math> with legs <math>x</math> and <math>500-x</math> and hypotenuse <math>400</math>. By the [[Pythagorean Theorem]],  
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 30: Line 33:
  
 
and applying the [[quadratic formula]] we get that  
 
and applying the [[quadratic formula]] we get that  
<math>x=250\pm 50\sqrt{7}</math>.  Since <math>BF > AE</math> we take the positive sign because and so our answer is <math>p+q+r = 250 + 50 + 7 = 307</math>.
+
<math>x=250\pm 50\sqrt{7}</math>.  Since <math>BF > AE,</math> we take the positive root, and our answer is <math>p+q+r = 250 + 50 + 7 = 307</math>.
 +
 
 +
=== 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 <math>\overline{AB}</math> be <math>M</math> and let <math>FB = x</math>, so then <math>MF = 450 - x</math> and <math>AF = 900 - x</math>. Drawing <math>\overline{AO}</math>, we have <math>\triangle OEF\sim\triangle AOF</math>, so
 +
<cmath>\frac{OF}{EF} = \frac{AF}{OF}\Rightarrow (OF)^2 = 400(900 - x).</cmath>
 +
By the Pythagorean Theorem on <math>\triangle OMF</math>,
 +
<cmath>(OF)^2 = 450^2 + (450 - x)^2.</cmath>
 +
Setting these two expressions for <math>(OF)^2</math> equal and solving for <math>x</math> (it is helpful to scale the problem down by a factor of 50 first), we get <math>x = 250\pm 50\sqrt{7}</math>. Since <math>BF > AE</math>, we want the value <math>x = 250 + 50\sqrt{7}</math>, and the answer is <math>250 + 50 + 7 = \boxed{307}</math>.
 +
 
 +
=== Solution 4 (Abusing Stewart) ===
 +
Let <math>x = BF</math>, so <math>AE = 500-x</math>. Let <math>a = OE</math>, <math>b = OF</math>. Applying Stewart's Theorem on triangles <math>AOB</math> twice, first using <math>E</math> as the base point and then <math>F</math>, we arrive at the equations <cmath>(450 \sqrt{2})^2 (900) = 900(500-x)(400+x) + a^2 (900)</cmath> and <cmath>(450 \sqrt{2})^2 (900) = 900x(900-x) + b^2 (900)</cmath> Now applying law of sines and law of cosines on <math>\triangle EOF</math> yields <cmath>\frac{1}{2} ab \sin 45^{\circ}  = \frac{4}{9} \times \frac{1}{4}  \times 900^2 = 202500</cmath> and <cmath>a^2+b^2- 2 ab \cos 45^{\circ} = 160000</cmath> Solving for <math>ab</math> from the sines equation and plugging into the law of cosines equation yields <math>a^2+b^2 = 290000</math>. We now finish by adding the two original stewart equations and obtaining: <cmath>2(450\sqrt{2})^2 = (500-x)(400+x)+x(900-x)+520000</cmath> This is a quadratic which only takes some patience to solve for <math>x = 250 + 50\sqrt{7}</math>
 +
 
 +
=== Solution 5 (Complex Numbers) ===
 +
Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with <math>o = 0, a = -450 + 450i, b = 450 + 450i</math>, and <math>f = x + 450i</math>. Since <math>EF</math> = 400, <math>e = (x-400) + 450i</math>. From <math>\angle{EOF} = 45^{\circ}</math>, we can deduce that the rotation of point <math>F</math> 45 degrees counterclockwise, <math>E</math>, and the origin are collinear. In other words, <cmath>\dfrac{e^{i \frac{\pi}{4}} \cdot (x + 450i)}{(x - 400) + 450i}</cmath> is a real number. Simplyfying using the fact that <math>e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}</math>, clearing the denominator, and setting the imaginary part equal to <math>0</math>, we eventually get the quadratic <cmath>x^2 - 400x + 22500 = 0</cmath> which has solutions <math>x = 200 \pm 50\sqrt{7}</math>. It is given that <math>AE < BF</math>, so <math>x = 200 - 50\sqrt{7}</math> and <cmath>BF = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7} \Rightarrow \boxed{307}.</cmath>
 +
 
 +
-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 <math>\triangle GOE \sim \triangle OHE</math>, <math>\frac{GO}{OE} = \frac{450}{x}</math>, and by [[Angle Bisector Theorem]], <math>\frac{GF}{FE} = \frac{450}{x}</math>. Thus, <math>GF = \frac{450 \cdot 400}{x}</math>. <math>AF = AH-FH = 50+x</math>, and <math>KA = EB</math> (90 degree rotation), and now we can bash on 2 similar triangles <math>\triangle GAK \sim \triangle GHO</math>.
 +
 
 +
<cmath>\frac{GA}{AK} = \frac{GH}{OH}</cmath>
 +
<cmath>\frac{\frac{450 \cdot 400}{x}-50-x}{450-x} = \frac{\frac{450 \cdot 400}{x}+400-x}{450}</cmath>
 +
I hope you like expanding
 +
<cmath>x^2 - 850x + \frac{81000000}{x} = -450x - 22500 + \frac{81000000}{x}</cmath>
 +
<cmath>x^2 - 400x + 22500 = 0</cmath>
 +
Quadratic formula gives us
 +
<cmath>x = 200 \pm 50 \sqrt{7}</cmath>
 +
Since AE < BF
 +
<cmath>x = 200 - 50 \sqrt{7}</cmath>
 +
Thus,
 +
<cmath>BF = 250 + 50 \sqrt{7}</cmath>
 +
So, our answer is <math>\boxed{307}</math>.
 +
 
 +
-AlexLikeMath
 +
 
 +
=== Solution 7 (Using a Circle) ===
 +
<center><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), H=(4.5-0.5*sqrt(7),4.5), I=(0,4.5), J=(4.5-0.5*sqrt(7),9);
 +
draw(A--B--C--D--A);
 +
draw(E--O--F);
 +
draw(J--G);
 +
draw(E--G--F);
 +
draw(G--H--O--G);
 +
draw(I--H);
 +
draw(circle(G,2*sqrt(2)));
 +
markscalefactor=0.05;
 +
draw(rightanglemark(E,G,F));
 +
dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^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("\(H\)",H,(-1,1));
 +
label("\(I\)",I,(-1,0));
 +
label("\(J\)",J,(0,1));
 +
label("\(O\)",O,(1,-1));
 +
</asy></center>
 +
 
 +
We know that G is on the perpendicular bisector of <math>EF</math>, which means that <math>EJ=JF=200</math>, <math>EG=GF=200\sqrt{2}</math> and <math>GH=250</math>. Now, let <math>HO</math> be equal to <math>x</math>. We can set up an equation with the Pythagorean Theorem:
 +
 
 +
<cmath>
 +
\begin{align*}
 +
\sqrt{x^2+250^2}&=(200\sqrt{2})^2 \\
 +
x^2+62500&=80000 \\
 +
x^2&=17500 \\
 +
x&=50\sqrt{7}
 +
\end{align*}
 +
</cmath>
 +
 
 +
Now, since <math>IO=450</math>,
 +
 
 +
<cmath>
 +
\begin{align*}
 +
HI&=450-x \\
 +
&=450-50\sqrt{7} \\
 +
\end{align*} \\
 +
</cmath>
 +
 
 +
Since <math>HI=AJ</math>, we now have:
 +
 
 +
<cmath>
 +
\begin{align*}
 +
BF&=AB-AJ-JF \\
 +
&=900-(450-50\sqrt{7})-200 \\
 +
&=250+50\sqrt{7} \\
 +
\end{align*}
 +
</cmath>
 +
 
 +
This means that our answer would be <math>250+50+7=\boxed{307}</math>
 +
 
 +
~Jerry_Guo
 +
 
 +
=== Solution 8 (More Similar Triangles) ===
 +
 +
Construct <math>BO, AO.</math> Let <math>\angle{FOB} = \alpha.</math> Also let <math>FB = x</math> then <math>AE = 500-x.</math> We then have from simple angle-chasing:
 +
<cmath>
 +
\begin{align*}
 +
\angle{BFO} = 135 - \alpha \\
 +
\angle{OFE} = 45 + \alpha \\
 +
\angle{EOA} = 45 - \alpha \\
 +
\angle{AEO} = 90 + \alpha \\
 +
\angle{OEF} = 90 - \alpha.
 +
\end{align*}
 +
</cmath>
 +
From AA similarity we have <cmath>\triangle{EOB} \sim \triangle{EFO}.</cmath> This gives the ratios,
 +
<cmath>\dfrac{400 + x}{EO} = \dfrac{450\sqrt{2}}{FO}.</cmath>
 +
Similarly from AA similarity <cmath>\triangle{FOA} \sim \triangle{FEO}.</cmath> So we get the ratios <cmath>\dfrac{EO}{450\sqrt{2}} = \dfrac{FO}{900-x}.</cmath> We can multiply to get
 +
<cmath>\dfrac{400 + x}{450\sqrt{2}} = \dfrac{450\sqrt{2}}{900 - x}.</cmath> Cross-multiplying reveals
 +
<cmath>360000 + 500x - x^2 = 405000.</cmath> Bringing everything to one side we have <cmath>x^2 - 500x + 45000 = 0.</cmath> By the quadratic formula we get <cmath>x = \dfrac{500 + \sqrt{500^2 - 4\cdot45000}}{2} = \dfrac{500 + \sqrt{70000}}{2} = \dfrac{500 + 100\sqrt{7}}{2} = 250 + 50\sqrt{7}.</cmath>
 +
Therefore
 +
<cmath>p+q+r = 250 + 50 + 7 = \boxed{307}.</cmath>
 +
~aa1024
 +
 
 +
 
 +
=== Solution 9 ===
 +
we use ratio lemma and stewarts theorem
 +
Connect OA,OE,OF,OB and let AE = x while BF = 500 - x
 +
let angle AOE = y, hence BOF = 45 - y
 +
now we apply stewarts 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
  
 
== See also ==
 
== See also ==
Line 36: Line 205:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 21:21, 11 February 2021

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), H=(4.5-0.5*sqrt(7),4.5), I=(0,4.5), J=(4.5-0.5*sqrt(7),9); draw(A--B--C--D--A); draw(E--O--F); draw(J--G); draw(E--G--F); draw(G--H--O--G); draw(I--H); draw(circle(G,2*sqrt(2))); markscalefactor=0.05; draw(rightanglemark(E,G,F)); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^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("\(H\)",H,(-1,1)); label("\(I\)",I,(-1,0)); label("\(J\)",J,(0,1)); label("\(O\)",O,(1,-1)); [/asy]

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{align*} \sqrt{x^2+250^2}&=(200\sqrt{2})^2 \\ x^2+62500&=80000 \\ x^2&=17500 \\ x&=50\sqrt{7} \end{align*}

Now, since $IO=450$,

\begin{align*} HI&=450-x \\ &=450-50\sqrt{7} \\ \end{align*} \\

Since $HI=AJ$, we now have:

\begin{align*} BF&=AB-AJ-JF \\ &=900-(450-50\sqrt{7})-200 \\ &=250+50\sqrt{7} \\ \end{align*}

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{align*} \angle{BFO} = 135 - \alpha \\ \angle{OFE} = 45 + \alpha \\ \angle{EOA} = 45 - \alpha \\ \angle{AEO} = 90 + \alpha \\ \angle{OEF} = 90 - \alpha. \end{align*} 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 stewarts theorem Connect OA,OE,OF,OB and let AE = x while BF = 500 - x let angle AOE = y, hence BOF = 45 - y now we apply stewarts 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

See also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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