Difference between revisions of "2018 AIME I Problems/Problem 2"

(Created page with "==Problem== What is the area of the polygon whose vertices are the points of intersection of the curves <math>x^2 + y^2 =25</math> and <math>(x-4)^2 + 9y^2 = 81 ?</math> ==S...")
 
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==Solution==
 
==Solution==
  
The first curve is a circle with radius <math>5</math> centered at the origin, and the second curve is an ellipse with center <math>(4,0)</math> and end points of <math>(-5,0)</math> and <math>(13,0)</math>. Finding points of intersection, we get <math>(-5,0)</math>, <math>(4,3)</math>, and <math>(4,-3)</math>, forming a triangle with height of <math>9</math> and base of <math>6.</math> So the area of this triangle is $9 \cdot 6 \cdot 0.5 =27 \
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The first curve is a circle with radius <math>5</math> centered at the origin, and the second curve is an ellipse with center <math>(4,0)</math> and end points of <math>(-5,0)</math> and <math>(13,0)</math>. Finding points of intersection, we get <math>(-5,0)</math>, <math>(4,3)</math>, and <math>(4,-3)</math>, forming a triangle with height of <math>9</math> and base of <math>6.</math> So the area of this triangle is $9 \cdot 6 \cdot 0.5 =27 \$

Revision as of 22:15, 28 February 2018

Problem

What is the area of the polygon whose vertices are the points of intersection of the curves $x^2 + y^2 =25$ and $(x-4)^2 + 9y^2 = 81 ?$

Solution

The first curve is a circle with radius $5$ centered at the origin, and the second curve is an ellipse with center $(4,0)$ and end points of $(-5,0)$ and $(13,0)$. Finding points of intersection, we get $(-5,0)$, $(4,3)$, and $(4,-3)$, forming a triangle with height of $9$ and base of $6.$ So the area of this triangle is $9 \cdot 6 \cdot 0.5 =27 $

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