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2006 AIME I Problems/Problem 1: Difference between revisions

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== Problem ==
== Problem ==
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 ==
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>.
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]]:


<div style="text-align:center"><math> (AD)^2 = (AC)^2 + (CD)^2 </math></div>
== Solution 1 ==


<div style="text-align:center"><math> (AC)^2 = (AB)^2 + (BC)^2 </math></div>
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><!--Asymptote by joml88-->
Using the [[Pythagorean Theorem]], we get the following two equations:
<cmath>AD^2 = AC^2 + CD^2</cmath>
<cmath>AC^2 = AB^2 + BC^2</cmath>
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{84}</math>.


Substituting <math>(AB)^2 + (BC)^2 </math> for <math> (AC)^2 </math>:
== See Also ==


<div style="text-align:center"><math> (AD)^2 = (AB)^2 + (BC)^2 + (CD)^2 </math></div>
Plugging in the given information:
<div style="text-align:center"><math> (AD)^2 = (18)^2 + (21)^2 + (14)^2 </math></div>
<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{84}</math>.
== See also ==
{{AIME box|year=2006|n=I|before=First Question|num-a=2}}
{{AIME box|year=2006|n=I|before=First Question|num-a=2}}
 
{{MAA Notice}}
[[Category:Intermediate Geometry Problems]]
[[Category:Introductory Geometry Problems]]

Latest revision as of 06:22, 30 August 2025

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 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] Using the Pythagorean Theorem, we get the following two equations: \[AD^2 = AC^2 + CD^2\] \[AC^2 = AB^2 + BC^2\] Substituting $AB^2 + BC^2$ for $AC^2$ gives us $AD^2 = AB^2 + BC^2 + CD^2$. Plugging in the given information, we get $AD^2 = 18^2 + 21^2 + 14^2 = 961 \implies AD = 31$, so the perimeter is $AB+BC+CD+AD = 18+21+14+31 = \boxed{84}$.

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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