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Difference between revisions of "2005 AIME I Problems/Problem 7"

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=== Solution 2 ===
 
=== Solution 2 ===
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<center>[[Image:AIME_2005I_Solution_7_2.png]]</center>
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Extend <math>AD</math> and <math>BC</math> to an intersection at point <math>E</math>. We get an [[equilateral triangle]] <math>ABE</math>. Solve <math>\triangle CDP</math> using the [[Law of Cosines]], denoting the length of a side of <math>\triangle ABE</math> as <math>s</math>. We get <math>12^2 = (s - 10)^2 + (s - 8)^2 - 2(s - 10)(s - 8)\cos 60 \Longrightarrow 144 = 2s^2 - 36s + 164 - (s^2 - 18s + 80)</math>. This boils down to a [[quadratic equation]]: <math>0 = s^2 - 18s + 60</math>; the [[quadratic formula]] yields the (discard the negative result) same result of <math>9 + \sqrt{141}</math>.
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Extend <math>AD</math> and <math>BC</math> to an intersection at point <math>E</math>. We get an [[equilateral triangle]] <math>ABE</math>. We denote the length of a side of <math>\triangle ABE</math> as <math>s</math> and solve for it using the [[Law of Cosines]]: <cmath>12^2 = (s - 10)^2 + (s - 8)^2 - 2(s - 10)(s - 8)\cos{60}</cmath>
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<cmath>144 = 2s^2 - 36s + 164 - (s^2 - 18s + 80)</cmath> This simplifies to <math>s^2 - 18s + 60=0</math>; the [[quadratic formula]] yields the (discard the negative result) same result of <math>9 + \sqrt{141}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 21:49, 9 November 2007

Problem

In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers, find $p+q.$

Solution

Solution 1

AIME 2005I Solution 7 1.png

Draw the perpendiculars from $C$ and $D$ to $AB$, labeling the intersection points as $E$ and $F$. This forms 2 $30-60-90$ right triangles, so $AE = 5$ and $BF = 4$. Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$, we find another right triangle $\triangle DGC$. $DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$. The Pythagorean theorem yields that $GC^2 = 12^2 - \sqrt{3}^2 = 141$, so $EF = GC = \sqrt{141}$. Therefore, $AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}$, and $p + q = 150$.

Solution 2

AIME 2005I Solution 7 2.png

Extend $AD$ and $BC$ to an intersection at point $E$. We get an equilateral triangle $ABE$. We denote the length of a side of $\triangle ABE$ as $s$ and solve for it using the Law of Cosines: \[12^2 = (s - 10)^2 + (s - 8)^2 - 2(s - 10)(s - 8)\cos{60}\] \[144 = 2s^2 - 36s + 164 - (s^2 - 18s + 80)\] This simplifies to $s^2 - 18s + 60=0$; the quadratic formula yields the (discard the negative result) same result of $9 + \sqrt{141}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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