# Difference between revisions of "1954 AHSME Problems/Problem 46"

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− | First we extend the line with <math>A</math> and the line <math>B</math> so that they both meet the line with <math>C</math>, forming an equilateral triangle. Let the vertices of this triangle be <math>D</math>, <math>E</math>, and <math>F</math>. We know it is equilateral because of the angle of <math>60^\circ</math> shown, and because the tangent lines <math>\overline{EF}</math> and <math>\overline{DE}</math> are congruent. <asy> | + | First we extend the line with <math>A</math> and the line with <math>B</math> so that they both meet the line with <math>C</math>, forming an equilateral triangle. Let the vertices of this triangle be <math>D</math>, <math>E</math>, and <math>F</math>. We know it is equilateral because of the angle of <math>60^\circ</math> shown, and because the tangent lines <math>\overline{EF}</math> and <math>\overline{DE}</math> are congruent. <asy> |

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## Latest revision as of 18:01, 24 May 2020

## Problem 46

In the diagram, if points and are points of tangency, then equals:

## Solution 1

First we extend the line with and the line with so that they both meet the line with , forming an equilateral triangle. Let the vertices of this triangle be , , and . We know it is equilateral because of the angle of shown, and because the tangent lines and are congruent. We can see, because , , and are points of tangency, that circle is inscribed in . The height of an equilateral triangle is exactly times the radius of a circle inscribed in it. Let the height of be . We can see that the radius of the circle equals . Thus Subtracting from gives us so our answer is .