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Difference between revisions of "2006 AIME I Problems/Problem 1"

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== Problem ==
 
== Problem ==
−
In [[convex polygon|convex]] [[hexagon]] <math>ABCDEF</math>, all six sides are congruent, <math>\angle A</math> and <math>\angle D</math> are [[right angle]]s, and <math>\angle B, \angle C, \angle E,</math> and <math>\angle F</math> are [[congruent]]. The area of the hexagonal region is <math>2116(\sqrt{2}+1).</math> Find <math>AB</math>.
+
In [[quadrilateral]] <math> ABCD</math>, <math>\angle B </math> is a [[right angle]], [[diagonal]] <math>\overline{AC}</math> is [[perpendicular]] to <math>\overline{CD}</math>, <math>AB=18</math>, <math>BC=21</math>, and <math>CD=14</math>. Find the [[perimeter]] of <math>ABCD</math>.
  
 
== Solution ==
 
== Solution ==
−
Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>.  
+
From the problem statement, we construct the following diagram:
 +
<center><asy>
 +
pointpen = black; pathpen = black + linewidth(0.65);
 +
pair C=(0,0), D=(0,-14),A=(-(961-196)^.5,0),B=IP(circle(C,21),circle(A,18));
 +
D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C); D(rightanglemark(A,C,D,40)); D(rightanglemark(A,B,C,40));
 +
</asy></center><!-- Asymptote replacement for Image:Aime06i.1.PNG by joml88 -->
 +
Using the [[Pythagorean Theorem]]:
  
−
[[Image:2006_I_AIME-1.png]]
+
<div style="text-align:center"><math> (AD)^2 = (AC)^2 + (CD)^2 </math></div>
  
−
The diagonal <math>BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}</math>. Then the areas of the triangles AFB and CDE in total are <math>\frac{x^2}{2}\cdot 2</math>,
+
<div style="text-align:center"><math> (AC)^2 = (AB)^2 + (BC)^2 </math></div>
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and the area of the rectangle BCEF equals <math>x\cdot x\sqrt{2}=x^2\sqrt{2}</math>
 
  
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Then we have to solve the equation
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Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>:
−
<div style="text-align:center;">
 
−
<math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>.
 
  
−
<math>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</math>
+
<div style="text-align:center"><math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math></div>
  
−
<math>2116=x^2</math>
+
Plugging in the given information:
  
−
<math>x=46</math></div>
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<div style="text-align:center"><math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math></div>
  
−
Therefore, <math>AB</math> is <math>046</math>.
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<div style="text-align:center"><math> (AD)^2 = 961 </math></div>
 +
 
 +
<div style="text-align:center"><math> (AD)= 31 </math></div>
 +
 
 +
So the perimeter is <math> 18+21+14+31=84 </math>, and the answer is <math>\boxed{084}</math>.
  
 
== See also ==
 
== See also ==
Line 26: Line 33:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:43, 2 February 2014

Problem

In quadrilateral $ABCD$, $\angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$, $AB=18$, $BC=21$, and $CD=14$. Find the perimeter of $ABCD$.

Solution

From the problem statement, we construct the following diagram:

[asy] pointpen = black; pathpen = black + linewidth(0.65); pair C=(0,0), D=(0,-14),A=(-(961-196)^.5,0),B=IP(circle(C,21),circle(A,18)); D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C); D(rightanglemark(A,C,D,40)); D(rightanglemark(A,B,C,40)); [/asy]

Using the Pythagorean Theorem:

$(AD)^2 = (AC)^2 + (CD)^2$
$(AC)^2 = (AB)^2 + (BC)^2$

Substituting $(AB)^2 + (BC)^2$ for $(AC)^2$:

$(AD)^2 = (AB)^2 + (BC)^2 + (CD)^2$

Plugging in the given information:

$(AD)^2 = (18)^2 + (21)^2 + (14)^2$
$(AD)^2 = 961$
$(AD)= 31$

So the perimeter is $18+21+14+31=84$, and the answer is $\boxed{084}$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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