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

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== Solution ==
 
== Solution ==
  
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Let the side length be called <math>x</math>.
+
From the problem statement, we construct the following diagram: <div style="text-align:center">[[Image:Aime06i.1.PNG]]</div>
−
[[Image:Diagram1.png]]
 
  
−
Then <math>AB=BC=CD=DE=EF=AF=x</math>.
+
Using the [[Pythagorean Theorem]]:
  
−
The diagonal <math>BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}</math>.
+
<div style="text-align:center"><math> (AD)^2 = (AC)^2 + (CD)^2 </math></div>
  
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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>
 
  
−
Then we have to solve the equation
+
Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>:
  
−
<math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>.
+
<div style="text-align:center"><math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math></div>
  
−
<math>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</math>
+
Plugging in the given information:
  
−
<math>2116=x^2</math>
+
<div style="text-align:center"><math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math></div>
  
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<math>x=46</math>
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<div style="text-align:center"><math> (AD)^2 = 961 </math></div>
  
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Therefore, AB is 46.
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<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>084</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:54, 25 September 2007

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:

Aime06i.1.PNG

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 $084$.

See also

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