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

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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 | + | 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 <math>9 \cdot 6 \cdot 0.5 =27 \textbf</math> |

## Revision as of 22:16, 28 February 2018

## Problem

What is the area of the polygon whose vertices are the points of intersection of the curves and

## Solution

The first curve is a circle with radius centered at the origin, and the second curve is an ellipse with center and end points of and . Finding points of intersection, we get , , and , forming a triangle with height of and base of So the area of this triangle is $9 \cdot 6 \cdot 0.5 =27 \textbf$ (Error compiling LaTeX. ! Extra }, or forgotten $.)