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

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== Solution ==
 
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=== Solution 1 ===
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Draw the [[perpendicular]]s from <math>C</math> and <math>D</math> to <math>AB</math>, labeling the intersection points as <math>E</math> and <math>F</math>. This forms 2 <math>30-60-90</math> [[right triangle]]s, so <math>AE = 5</math> and <math>BF = 4</math>. Also, if we draw the horizontal line extending from <math>C</math> to a point <math>G</math> on the line <math>DE</math>, we find another right triangle. <math>DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}</math>. The [[Pythagorean theorem]] yields that <math>GC^2 = 12^2 - \sqrt{3}^2 = 141</math>, so <math>EF = GC = \sqrt{141}</math>. Therefore, <math>AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}</math>, and <math>p + q = 150</math>.  
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Draw the [[perpendicular]]s from <math>C</math> and <math>D</math> to <math>AB</math>, labeling the intersection points as <math>E</math> and <math>F</math>. This forms 2 <math>30-60-90</math> [[right triangle]]s, so <math>AE = 5</math> and <math>BF = 4</math>. Also, if we draw the horizontal line extending from <math>C</math> to a point <math>G</math> on the line <math>DE</math>, we find another right triangle <math>\triangle DGC</math>. <math>DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}</math>. The [[Pythagorean theorem]] yields that <math>GC^2 = 12^2 - \sqrt{3}^2 = 141</math>, so <math>EF = GC = \sqrt{141}</math>. Therefore, <math>AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}</math>, and <math>p + q = 150</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 21:28, 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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Extend $AD$ and $BC$ to an intersection at point $E$. We get an equilateral triangle $ABE$. Solve $\triangle CDP$ using the Law of Cosines, denoting the length of a side of $\triangle ABE$ as $s$. We get $12^2 = (s - 10)^2 + (s - 8)^2 - 2(s - 10)(s - 8)\cos 60 \Longrightarrow 144 = 2s^2 - 36s + 164 - (s^2 - 18s + 80)$. This boils down to a quadratic equation: $0 = s^2 - 18s + 60$; 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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