Difference between revisions of "2003 AIME II Problems/Problem 7"
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== Problem == | == Problem == | ||
− | Find the area of rhombus <math>ABCD</math> given that the | + | 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 == | ||
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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>. | ||
+ | |||
+ | ==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 == |
Latest revision as of 09:56, 16 September 2022
Problem
Find the area of rhombus given that the circumradii of 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, Since , , so Hence . Solving , . Thus The height of the rhombus is , so we want
~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.