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Difference between revisions of "2003 AIME II Problems/Problem 7"
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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>.
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==Solution 2==
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Let <math>\theta=\angle BDA</math>. Let <math>AB=BC=CD=x</math>. By the extended law of sines,
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<cmath>\frac{x}{\sin\theta}=25</cmath>
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Since <math>AC\perp BD</math>, <math>\angle CAD=90-\theta</math>, so
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<cmath>\frac{x}{\sin(90-\theta)=\cos\theta}=50</cmath>
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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
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<cmath>x=25\frac{2}{\sqrt{5}}\implies x^2=500</cmath>
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The height of the rhombus is <math>x\sin(2\theta)=2x\sin\theta\cos\theta</math>, so we want
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<cmath>2x^2\sin\theta\cos\theta=\boxed{400}</cmath>
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 +
~yofro
  
 
== See also ==
 
== See also ==

Revision as of 18:55, 15 August 2021

Problem

Find the area of rhombus $ABCD$ given that the radii of the circles circumscribed around 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

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

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