Difference between revisions of "2006 AIME I Problems/Problem 1"
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<cmath>AD^2 = AC^2 + CD^2</cmath> | <cmath>AD^2 = AC^2 + CD^2</cmath> | ||
<cmath>AC^2 = AB^2 + BC^2</cmath> | <cmath>AC^2 = AB^2 + BC^2</cmath> | ||
| − | Substituting <math>AB^2 + BC^2</math> for <math>AC^2</math> gives us | + | Substituting <math>AB^2 + BC^2</math> for <math>AC^2</math> gives us <math>AD^2 = AB^2 + BC^2 + CD^2</math>. Plugging in the given information, we get <math>AD^2 = 18^2 + 21^2 + 14^2 = 961 \implies AD = 31</math>, so the perimeter is <math>AB+BC+CD+AD = 18+21+14+31 = \boxed{084}</math>. |
| − | < | ||
| − | Plugging in the given information | ||
| − | < | ||
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== See Also == | == See Also == | ||
Latest revision as of 15:10, 3 February 2025
Problem
In quadrilateral 
, 
is a right angle, diagonal 
is perpendicular to 
, 
, 
, and 
. Find the perimeter of .
Solution 1
We construct the following diagram:
![[asy] pathpen = black; pair C=(0,0),D=(0,-14),A=(-sqrt(765),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/e/0/6/e065a4745d2a144435bbcb457cecd1e77fc50554.png)
Using the Pythagorean Theorem, we get the following two equations:
![\[AD^2 = AC^2 + CD^2\]](http://latex.artofproblemsolving.com/e/0/3/e03e010c17d0e007e3fdd0fb6b0af337f6e6d27e.png)
![\[AC^2 = AB^2 + BC^2\]](http://latex.artofproblemsolving.com/f/f/6/ff646ee41223da2ea3f0c44c30628991d9d981fe.png)
Substituting 
for 
gives us 
. Plugging in the given information, we get 
, so the perimeter 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. 
