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
(→‎Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
{{image}}
+
<center>[[Image:AIME_2005II_Solution_12_1.png]]</center>
−
Draw the perpendicular from <math>AB \perp OP</math>, with the intersection at <math>G</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>\displaystyle \angle EOG = \frac{x}{450}</math>, and of <math>\tan \angle FOG = \frac{y}{450}</math>.
+
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>.
  
−
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>\displaystyle \tan 45 = 1</math>, <math>1 - \frac{xy}{450^2} = \frac{x + y}{450}</math>. We know that <math>x + y = 400 \displaystyle</math>, so we can substitute this to find that <math>1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2</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 <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>.
  
−
A second equation can be set up using <math>x + y = 400 \displaystyle</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>.
+
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>.
  
 
Now, <math>BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}</math>. The answer is <math>250 + 50 + 7 = 307</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>.

Revision as of 21:03, 9 November 2007

Problem

Square $\displaystyle ABCD$ 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

AIME 2005II Solution 12 1.png

Let $G$ be a point on $AB$ such that $AB\perp OG$. Denote $x = EG$ and $y = FG$, and $x > y$ (since $AE < BF$ and $AG = BG$). The tangent of $\angle EOG = \frac{x}{450}$, and of $\tan \angle FOG = \frac{y}{450}$.

By the tangent addition rule $\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)$, we see that $\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \frac{y}{450}}$. Since $\tan 45 = 1$, $1 - \frac{xy}{450^2} = \frac{x + y}{450}$. We know that $x + y = 400$, so we can substitute this to find that $1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2$.

A second equation can be set up using $x + y = 400$. To solve for $y$, $x = 400 - y \Longrightarrow (400 - y)y = 150^2$. 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 $250 + 50 + 7 = 307$.

Solution 2

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 $\angle BOE+\angle AOE=45^\circ$ 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 $\displaystyle FG=400$. Now we have a right $\triangle BFG$ with legs $x$ and $500-x$ and hypotenuse 400. Then by the Pythagorean Theorem],

$\displaystyle(500-x)^2+x^2=400^2$

$250000-1000x+2x^2=16000$


$\displaystyle 90000-1000x+2x^2=0$

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 $\displaystyle p+q+r = 250 + 50 + 7 = 307$.

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=19296"