Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AIME II Problems/Problem 7"

(→‎Solution)
m (→‎Problem)
 
(12 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 +
__TOC__
 +
 
== Problem ==
 
== Problem ==
−
Find the area of rhombus <math>ABCD</math> given that the radii of the circles circumscribed around triangles <math>ABD</math> and <math>ACD</math> are <math>12.5</math> and <math>25</math>, respectively.
+
Find the area of rhombus <math>ABCD</math> given that the circumradii of triangles <math>ABD</math> and <math>ACD</math> are <math>12.5</math> and <math>25</math>, respectively.
  
 
== Solution ==
 
== Solution ==
−
The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD a and half of diagonal AC b. The length of the four sides of the rhombus is <math>\sqrt{a^2+b^2}</math>. The area of any triangle can be expressed as <math>a\cdot b\cdot c/4R</math>, where a, b, and c are the sides and R is the circumradius. Thus, the area of triangle ABD is <math>ab=2a(a^2+b^2)/(4\cdot12.5)</math>. Also, the area of triangle ABC is <math>ab=2b(a^2+b^2)/(4\cdot25)</math>. Setting these two expressions equal to each other and simplifying gives b=2a. Substitution yields a=10 and b=20, so the area of the rhombus is <math>20\cdot40/2=400</math>.
+
The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD <math>a</math> and half of diagonal AC <math>b</math>. The length of the four sides of the rhombus is <math>\sqrt{a^2+b^2}</math>.  
 +
 
 +
The area of any triangle can be expressed as <math>\frac{a\cdot b\cdot c}{4R}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are the sides and <math>R</math> is the circumradius. Thus, the area of <math>\triangle ABD</math> is <math>ab=2a(a^2+b^2)/(4\cdot12.5)</math>. Also, the area of <math>\triangle ABC</math> is <math>ab=2b(a^2+b^2)/(4\cdot25)</math>. Setting these two expressions equal to each other and simplifying gives <math>b=2a</math>. Substitution yields <math>a=10</math> and <math>b=20</math>, so the area of the rhombus is <math>20\cdot40/2=\boxed{400}</math>.
 +
 
 +
==Solution 2==
 +
Let <math>\theta=\angle BDA</math>. Let <math>AB=BC=CD=x</math>. By the extended law of sines,
 +
<cmath>\frac{x}{\sin\theta}=25</cmath>
 +
Since <math>AC\perp BD</math>, <math>\angle CAD=90-\theta</math>, so
 +
<cmath>\frac{x}{\sin(90-\theta)=\cos\theta}=50</cmath>
 +
Hence <math>x=25\sin\theta=50\cos\theta</math>. Solving <math>\tan\theta=2</math>, <math>\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\sqrt{5}}</math>. Thus
 +
<cmath>x=25\frac{2}{\sqrt{5}}\implies x^2=500</cmath>
 +
The height of the rhombus is <math>x\sin(2\theta)=2x\sin\theta\cos\theta</math>, so we want
 +
<cmath>2x^2\sin\theta\cos\theta=\boxed{400}</cmath>
 +
 
 +
~yofro
 +
 
 +
==Video Solution by Sal Khan==
 +
https://www.youtube.com/watch?v=jpKjXtywTlQ&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=16 - AMBRIGGS
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2003|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2003|n=II|num-b=6|num-a=8}}
 +
 +
[[Category: Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 08:56, 16 September 2022

Problem

Find the area of rhombus $ABCD$ given that the circumradii of triangles $ABD$ and $ACD$ are $12.5$ and $25$, respectively.

Solution

The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD $a$ and half of diagonal AC $b$. The length of the four sides of the rhombus is $\sqrt{a^2+b^2}$.

The area of any triangle can be expressed as $\frac{a\cdot b\cdot c}{4R}$, where $a$, $b$, and $c$ are the sides and $R$ is the circumradius. Thus, the area of $\triangle ABD$ is $ab=2a(a^2+b^2)/(4\cdot12.5)$. Also, the area of $\triangle ABC$ is $ab=2b(a^2+b^2)/(4\cdot25)$. Setting these two expressions equal to each other and simplifying gives $b=2a$. Substitution yields $a=10$ and $b=20$, so the area of the rhombus is $20\cdot40/2=\boxed{400}$.

Solution 2

Let $\theta=\angle BDA$. Let $AB=BC=CD=x$. By the extended law of sines, \[\frac{x}{\sin\theta}=25\] Since $AC\perp BD$, $\angle CAD=90-\theta$, so \[\frac{x}{\sin(90-\theta)=\cos\theta}=50\] Hence $x=25\sin\theta=50\cos\theta$. Solving $\tan\theta=2$, $\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\sqrt{5}}$. Thus \[x=25\frac{2}{\sqrt{5}}\implies x^2=500\] The height of the rhombus is $x\sin(2\theta)=2x\sin\theta\cos\theta$, so we want \[2x^2\sin\theta\cos\theta=\boxed{400}\]

~yofro

Video Solution by Sal Khan

https://www.youtube.com/watch?v=jpKjXtywTlQ&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=16 - AMBRIGGS

See also

2003 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_7&oldid=178247"