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

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<math>\angle CEF = 60^{\circ}</math> and triangle <math>AEF</math> is an equilateral triangle with side length 8. Thus, <math>AF = 8</math>.  
 
<math>\angle CEF = 60^{\circ}</math> and triangle <math>AEF</math> is an equilateral triangle with side length 8. Thus, <math>AF = 8</math>.  
  
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<math>ED = 10 - 8 = 2</math>. By the Law of Cosines, <math>EC^2 = ED^2 + DC^2 + 2 \cdot 12 \cdot \cos{\angle DEC}</math>. Thus, <math>144 = 4 + c^2 - 2c</math> and <math>c = 1 + \sqrt{141}</math>. <math>EC = FB</math>.  
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<math>ED = 10 - 8 = 2</math>. By the Law of Cosines, <math>CD^2 = ED^2 + EC^2 - 2 \cdot {ED} \cdot {EC} \cdot \cos{\angle DEC}</math>. Thus, <math>144 = 4 + c^2 - 2c</math> and <math>c = 1 + \sqrt{141}</math>. <math>EC = FB</math>.  
  
 
<math>AB = AF + FB = 8 + 1 + \sqrt{141} = 9 + \sqrt{141}</math> so the answer is <math>p+q=150</math>.
 
<math>AB = AF + FB = 8 + 1 + \sqrt{141} = 9 + \sqrt{141}</math> so the answer is <math>p+q=150</math>.
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-unhappyfarmer
  
 
== See also ==
 
== See also ==

Latest revision as of 23:43, 24 March 2025

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

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

Solution 3

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 $9 + \sqrt{141}$.

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 $p+q=150$.

Solution 5 (EASY EASY VERY TREMENDOUSLY EASY)

Draw a line from point $C$ to a new point $E$ on $AD$ parallel to $AB$. Draw a line from point $E$ to a new point $F$ on $AB$ parallel to $CD$. This creates parallelogram $CEFB$. 

 The reasoning for this is to connect the two angles that are congruent.

$\angle CEF = 60^{\circ}$ and triangle $AEF$ is an equilateral triangle with side length 8. Thus, $AF = 8$.

$ED = 10 - 8 = 2$. By the Law of Cosines, $CD^2 = ED^2 + EC^2 - 2 \cdot {ED} \cdot {EC} \cdot \cos{\angle DEC}$. Thus, $144 = 4 + c^2 - 2c$ and $c = 1 + \sqrt{141}$. $EC = FB$.

$AB = AF + FB = 8 + 1 + \sqrt{141} = 9 + \sqrt{141}$ so the answer is $p+q=150$.

-unhappyfarmer

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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