2003 AIME II Problems/Problem 7: Difference between revisions
Hashtagmath (talk | contribs) No edit summary |
|||
| Line 8: | Line 8: | ||
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>. | 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 | |||
== See also == | == See also == | ||
Revision as of 18:55, 15 August 2021
Problem
Find the area of rhombus 
given that the radii of the circles circumscribed around triangles 
and 
are 
and 
, respectively.
Solution
The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD 
and half of diagonal AC 
. The length of the four sides of the rhombus is 
.
The area of any triangle can be expressed as 
, where , , and 
are the sides and 
is the circumradius. Thus, the area of 
is 
. Also, the area of 
is 
. Setting these two expressions equal to each other and simplifying gives 
. Substitution yields 
and 
, so the area of the rhombus is 
.
Solution 2
Let 
. Let 
. By the extended law of sines,
![\[\frac{x}{\sin\theta}=25\]](http://latex.artofproblemsolving.com/e/2/8/e281210aebab701fb404bbc6aff08c0335a86b24.png)
Since 
, 
, so
![\[\frac{x}{\sin(90-\theta)=\cos\theta}=50\]](http://latex.artofproblemsolving.com/3/4/f/34ff0ef416c65ad450251ca8bf6050b99f8ba4b6.png)
Hence 
. Solving 
, 
. Thus
![\[x=25\frac{2}{\sqrt{5}}\implies x^2=500\]](http://latex.artofproblemsolving.com/c/a/9/ca900f3e85c2d4bc5599ec5cf9b9e50d756eec0a.png)
The height of the rhombus is 
, so we want
![\[2x^2\sin\theta\cos\theta=\boxed{400}\]](http://latex.artofproblemsolving.com/4/a/b/4ab6ec6da72e8309592e2864f1d8044e4196fce3.png)
~yofro
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
