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

Latest revision as of 19:20, 3 March 2020

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

$[asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label("$A$",(0,0),SW); label("$B$",(20.87,0),SE); label("$E$",(15.87,8.66),NE); label("$D$",(5,8.66),NW); label("$P$",(5,0),S); label("$Q$",(15.87,0),S); label("$C$",(16.87,7),E); label("$12$",(10.935,7.794),S); label("$10$",(2.5,4.5),W); label("$10$",(18.37,4.5),E); [/asy]$

Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$ . Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$ , and $BC=8$ and $EC=2$ . We are given that $DC=12$ . Since $\angle CED = 120^{\circ}$ , using Law of Cosines on $\bigtriangleup CED$ gives $12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})$ which gives $144-4=DE^2+2DE$ . Adding $1$ to both sides gives $141=(DE+1)^2$ , so $DE=\sqrt{141}-1$ . $\bigtriangleup DAP$ and $\bigtriangleup EBQ$ are both $30-60-90$ , so $AP=5$ and $BQ=5$ . $PQ=DE$ , and therefore $AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}$ .

Solution 2

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 $p + q = \boxed{150}$ .

Solution 3

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 $9 + \sqrt{141}$ .

Solution 4

Extend $BC$ and $AD$ to meet at point $E$ , forming an equilateral triangle $\triangle ABE$ . Draw a line from $C$ parallel to $AB$ so that it intersects $AD$ at point $F$ . Then, apply Stewart's Theorem on $\triangle CFE$ . Let $CE=x$ . $2x(x-2) + 12^2x = 2x^2 + x^2(x-2)$ $x^3 - 2x^2 - 140x = 0$ By the quadratic formula (discarding the negative result), $x = 1 + \sqrt{141}$ , giving $AB = 9 + \sqrt{141}$ for a final answer of $p+q=150$ .

@@ Line 20: / Line 20: @@
 label("$10$",(18.37,4.5),E);
 </asy>
+Draw line segment <math>DE</math> such that line <math>DE</math> is concurrent with line <math>BC</math>. Then, <math>ABED</math> is an isosceles trapezoid so <math>AD=BE=10</math>, and <math>BC=8</math> and <math>EC=2</math>. We are given that <math>DC=12</math>. Since <math>\angle CED = 120^{\circ}</math>, using Law of Cosines on <math>\bigtriangleup CED</math> gives <cmath>12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})</cmath> which gives <cmath>144-4=DE^2+2DE</cmath>. Adding <math>1</math> to both sides gives <math>141=(DE+1)^2</math>, so <math>DE=\sqrt{141}-1</math>. <math>\bigtriangleup DAP</math> and <math>\bigtriangleup EBQ</math> are both <math>30-60-90</math>, so <math>AP=5</math> and <math>BQ=5</math>. <math>PQ=DE</math>, and therefore <math>AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}</math>.
 === Solution 2 ===
 <center>[[Image:AIME_2005I_Solution_7_1.png]]</center>

2005 AIME I (Problems • Answer Key • Resources)
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