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

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

@@ Line 1: / Line 1: @@
+__TOC__
 == Problem ==
+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 <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 ==
-* [[2003 AIME II Problems/Problem 6| Previous problem]]
+{{AIME box|year=2003|n=II|num-b=6|num-a=8}}
-* [[2003 AIME II Problems/Problem 8| Next problem]]
-* [[2003 AIME II Problems]]
+[[Category: Intermediate Geometry Problems]]
+{{MAA Notice}}

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