Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (toc shift, fmt)
(format asy etc)
Line 4: Line 4:
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
−
<center>[[Image:AIME_2005II_Solution_12_1.png]]</center>
+
<center><asy>
−
Let <math>G</math> be a point on <math>AB</math> such that <math>AB\perp OG</math>. Denote <math>x = EG</math> and <math>y = FG</math>, and <math>x > y</math> (since <math>AE < BF</math> and <math>AG = BG</math>). The tangent of <math>\angle EOG = \frac{x}{450}</math>, and of <math>\tan \angle FOG = \frac{y}{450}</math>.
+
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 -->
  
−
By the [[trigonometric identity|tangent addition rule]] <math>\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)</math>, we see that <math>\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \frac{y}{450}}</math>. Since <math>\tan 45 = 1</math>, <math>1 - \frac{xy}{450^2} = \frac{x + y}{450}</math>. We know that <math>x + y = 400</math>, so we can substitute this to find that <math>1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2</math>.
+
Let <math>G</math> be the foot of the [[perpendicular]] from <math>O</math> to <math>AB</math>. Denote <math>x = EG</math> and <math>y = FG</math>, and <math>x > y</math> (since <math>AE < BF</math> and <math>AG = BG</math>). Then <math>\tan \angle EOG = \frac{x}{450}</math>, and <math>\tan \angle FOG = \frac{y}{450}</math>.
  
−
A second equation can be set up using <math>x + y = 400</math>. To solve for <math>y</math>, <math>x = 400 - y \Longrightarrow (400 - y)y = 150^2</math>. This is a quadratic with roots <math>200 \pm 50\sqrt{7}</math>. Since <math>y < x</math>, use the smaller root, <math>200 - 50\sqrt{7}</math>.
+
By the [[trigonometric identity|tangent addition rule]] <math>\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)</math>, we see that <cmath>\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.</cmath> Since <math>\tan 45 = 1</math>, this simplifies to <math>1 - \frac{xy}{450^2} = \frac{x + y}{450}</math>. We know that <math>x + y = 400</math>, so we can substitute this to find that <math>1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2</math>.
  
−
Now, <math>BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}</math>. The answer is <math>250 + 50 + 7 = 307</math>.
+
Substituting <math>x = 400 - y</math> again, we know have <math>xy = (400 - y)y = 150^2</math>. This is a quadratic with roots <math>200 \pm 50\sqrt{7}</math>. Since <math>y < x</math>, use the smaller root, <math>200 - 50\sqrt{7}</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 ===
−
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 BOE+\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
+
<center><asy>
−
<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 400. Then by the [[Pythagorean Theorem]],  
+
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>
−
<math>(500-x)^2+x^2=400^2</math>
 
  
−
<math> 250000-1000x+2x^2=16000</math>
+
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>90000-1000x+2x^2=0</math>
+
<cmath>\begin{align*}
 +
(500-x)^2+x^2&=400^2 \\
 +
250000-1000x+2x^2&=16000 \\
 +
90000-1000x+2x^2&=0 \end{align*}</cmath>
  
 
and applying the [[quadratic formula]] we get that  
 
and applying the [[quadratic formula]] we get that  

Revision as of 10:33, 27 June 2008

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 $p+q+r.$

Solution

Solution 1

[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

[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), 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 and 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 $400$. Then 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 sign because and so our answer is $p+q+r = 250 + 50 + 7 = 307$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_12&oldid=26788"