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

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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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<asy>draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle);
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draw((5,8.66)--(5,0));
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draw((15.87,8.66)--(15.87,0));
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draw((5,8.66)--(16.87,6.928));
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label("$A$",(0,0),SW);
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label("$B$",(20.87,0),SE);
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label("$E$",(15.87,8.66),NE);
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label("$D$",(5,8.66),NW);
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label("$P$",(5,0),S);
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label("$Q$",(15.87,0),S);
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label("$C$",(16.87,7),E);
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label("$12$",(10.935,7.794),S);
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label("$10$",(2.5,4.5),W);
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label("$10$",(18.37,4.5),E);
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</asy>
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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>.
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 +
=== Solution 2 ===
 
<center>[[Image:AIME_2005I_Solution_7_1.png]]</center>
 
<center>[[Image:AIME_2005I_Solution_7_1.png]]</center>
  
 
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 = \boxed{150}</math>.
 
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 = \boxed{150}</math>.
  
−
=== Solution 2 ===
+
=== Solution 3 ===
 
<center>[[Image:AIME_2005I_Solution_7_2.png]]</center>
 
<center>[[Image:AIME_2005I_Solution_7_2.png]]</center>
 
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>
 
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>
 
<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>.
 
<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>.
  
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=== Solution 3 ===
+
=== Solution 4 ===
 
Extend <math>BC</math> and <math>AD</math> to meet at point <math>E</math>, forming an equilateral triangle <math>\triangle ABE</math>. Draw a line from <math>C</math> parallel to <math>AB</math> so that it intersects <math>AD</math> at point <math>F</math>. Then, apply [[Stewart's Theorem]] on <math>\triangle CFE</math>. Let <math>CE=x</math>. <cmath>2x(x-2) + 12^2x = 2x^2 + x^2(x-2)</cmath> <cmath>x^3 - 2x^2 - 140x = 0</cmath> By the quadratic formula (discarding the negative result), <math>x = 1 + \sqrt{141}</math>, giving <math>AB = 9 + \sqrt{141}</math> for a final answer of <math>p+q=150</math>.
 
Extend <math>BC</math> and <math>AD</math> to meet at point <math>E</math>, forming an equilateral triangle <math>\triangle ABE</math>. Draw a line from <math>C</math> parallel to <math>AB</math> so that it intersects <math>AD</math> at point <math>F</math>. Then, apply [[Stewart's Theorem]] on <math>\triangle CFE</math>. Let <math>CE=x</math>. <cmath>2x(x-2) + 12^2x = 2x^2 + x^2(x-2)</cmath> <cmath>x^3 - 2x^2 - 140x = 0</cmath> By the quadratic formula (discarding the negative result), <math>x = 1 + \sqrt{141}</math>, giving <math>AB = 9 + \sqrt{141}</math> for a final answer of <math>p+q=150</math>.
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 +
=== Solution 5 (EASY EASY VERY TREMENDOUSLY EASY) ===
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 +
Draw a line from point <math>C</math> to a new point <math>E</math> on <math>AD</math> parallel to <math>AB</math>. Draw a line from point <math>E</math> to a new point <math>F</math> on <math>AB</math> parallel to <math>CD</math>. This creates parallelogram <math>CEFB</math>.
 +
  The reasoning for this is to connect the two angles that are congruent.
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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>.
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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>.
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 +
<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 ==
Line 21: Line 53:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

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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