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

Revision as of 15:49, 4 March 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

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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. $DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$ . 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

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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 a quadratic equation: $0 = s^2 - 18s + 60$ ; the quadratic formula yields the (discard the negative result) same result of $9 + \sqrt{141}$ .

@@ Line 1: / Line 1: @@
 == Problem ==
-In quadrilateral <math> ABCD, BC=8, CD=12, AD=10, </math> and <math> m\angle A= m\angle B = 60^\circ. </math> Given that <math> AB = p + \sqrt{q}, </math> where <math> p </math> and <math> q </math> are positive integers, find <math> p+q. </math>
+In [[quadrilateral]] <math> ABCD,\ BC=8,\ CD=12,\ AD=10, </math> and <math> m\angle A= m\angle B = 60^\circ. </math> Given that <math> AB = p + \sqrt{q}, </math> where <math> p </math> and <math> q </math> are [[positive]] [[integer]]s, find <math> p+q. </math>
+__TOC__
 == Solution ==
+=== Solution 1 ===
+{{image}}
+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>.
+=== Solution 2 ===
+{{image}}
+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 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>.
 == See also ==
-* [[2005 AIME I Problems]]
+{{AIME box|year=2005|n=I|num-b=6|num-a=8}}
+[[Category:Intermediate Geometry Problems]]

2005 AIME I (Problems • Answer Key • Resources)
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