Difference between revisions of "2005 AIME II Problems/Problem 12"
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== Problem == | == Problem == | ||
− | [[Square]] <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__ | |
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
<center>[[Image:AIME_2005II_Solution_12_1.png]]</center> | <center>[[Image:AIME_2005II_Solution_12_1.png]]</center> | ||
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>. | 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>. | ||
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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>. | ||
− | == 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 | 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 | ||
− | <math>\angle{FOG}=45^\circ</math> too and by SAS we know that <math>\triangle FOE\cong \triangle FOG</math> so <math> | + | <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]], |
− | <math> | + | <math>(500-x)^2+x^2=400^2</math> |
<math> 250000-1000x+2x^2=16000</math> | <math> 250000-1000x+2x^2=16000</math> | ||
− | + | <math>90000-1000x+2x^2=0</math> | |
− | <math> | ||
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> | + | <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>. |
== See also == | == See also == |
Revision as of 10:44, 10 November 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
Solution 1
![AIME 2005II Solution 12 1.png](https://wiki-images.artofproblemsolving.com//c/cf/AIME_2005II_Solution_12_1.png)
Let be a point on
such that
. 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
.
Solution 2
Label , so
. Rotate
about
until
lies on
. Now we know that
therefore
also since
is the center of the square. Label the new triangle that we created
. Now we know that rotation preserves angles and side lengths, so
and
. Draw
and
. Notice that
since rotations preserve the same angles so
too and by SAS we know that
so
. Now we have a right
with legs
and
and hypotenuse 400. Then by the Pythagorean Theorem,
and applying the quadratic formula we get that
. Since
we take the positive sign because and so our 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 |