Difference between revisions of "2005 AIME II Problems/Problem 12"
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== Problem == | == Problem == | ||
− | + | [[Square]] <math>\displaystyle 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 | ||
== Solution == | == Solution == | ||
+ | {{image}} | ||
+ | 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>. | ||
− | {{ | + | 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>. |
− | == | + | 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>. |
− | + | Now, <math>BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}</math>. The answer is <math>250 + 50 + 7 = 307</math>. | |
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+ | == See also == | ||
+ | {{AIME box|year=2005|n=II|num-b=11|num-a=13}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 18:20, 22 March 2007
Problem
Square has center
and
are on
with
and
between
and
and
Given that
where
and
are positive integers and
is not divisible by the square of any prime, find
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Draw the perpendicular from , with the intersection at
. Denote
and
, and
(since
and
). The tangent of
, and of
.
By the tangent addition rule , we see that
. Since
,
. We know that
, so we can substitute this to find that
.
A second equation can be set up using . To solve for
,
. This is a quadratic with roots
. Since
, use the smaller root,
.
Now, . The answer is
.
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 |