Difference between revisions of "2006 AIME I Problems/Problem 1"
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== Problem == | == Problem == | ||
| − | In quadrilateral <math> ABCD , \angle B </math> is a right angle, diagonal <math> \overline{AC} </math> is perpendicular to <math> \overline{CD}, | + | 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 == | ||
| − | Using the Pythagorean Theorem: | + | 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]]: | ||
| − | <math> (AD)^2 = (AC)^2 + (CD)^2 </math> | + | <div style="text-align:center"><math> (AD)^2 = (AC)^2 + (CD)^2 </math></div> |
| − | <math> (AC)^2 = (AB)^2 + (BC)^2 </math> | + | <div style="text-align:center"><math> (AC)^2 = (AB)^2 + (BC)^2 </math></div> |
Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>: | Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>: | ||
| − | <math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math> | + | <div style="text-align:center"><math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math></div> |
Plugging in the given information: | Plugging in the given information: | ||
| − | <math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math> | + | <div style="text-align:center"><math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math></div> |
| − | <math> (AD)^2 = 961 </math> | + | <div style="text-align:center"><math> (AD)^2 = 961 </math></div> |
| − | <math> (AD)= 31 </math> | + | <div style="text-align:center"><math> (AD)= 31 </math></div> |
| − | So the perimeter is | + | So the perimeter is <math> 18+21+14+31=84 </math>, and the answer is <math>\boxed{084}</math>. |
| − | <math> 18+21+14+31=084 </math> | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2006|n=I|before=First Question|num-a=2}} | |
| + | |||
| + | [[Category:Intermediate Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:43, 2 February 2014
Problem
In quadrilateral 
, 
is a right angle, diagonal 
is perpendicular to 
, 
, 
, and 
. Find the perimeter of .
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]](http://latex.artofproblemsolving.com/f/e/5/fe5ba5b6cc93f6a188115c7f4281a190091a9844.png)
Using the Pythagorean Theorem:
Substituting 
for 
:
Plugging in the given information:
So the perimeter is 
, and the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
